Chapter 10 Hypothesis Tests Using a Single Sample

Chapter 10

Hypothesis Tests Using a Single Sample

2 Copyright (c) 2001 Brooks/Cole, a division of Thomson Learning, Inc.

BASICS

In statistics, a hypothesis is a statement about a population characteristic.


FORMAL STRUCTURE

Hypothesis Tests are based on an reductio ad absurdum form of argument.

Specifically, we make an assumption and then attempt to show that assumption leads to an absurdity or contradiction, hence the assumption is wrong.


FORMAL STRUCTURE

The null hypothesis, denoted H0 is a statement or claim about a population characteristic that is initially assumed to be true.

The null hypothesis is so named because it is the “starting point” for the investigation. The phrase “there is no difference” is often used in its interpretation.


FORMAL STRUCTURE

The alternate hypothesis, denoted by Ha is the competing claim.

The is a statement about the same population characteristic that is used in the null hypothesis. Generally, alternate hypothesis is a statement that specifies that the population has a value different, in some way, from the value given in the null hypothesis.


FORMAL STRUCTURE

Rejection of the null hypothesis will imply the acceptance of this alternative hypothesis.

Assume H0 is true and attempt to show this leads to an absurdity, hence H0 is false and Ha is true.


FORMAL STRUCTURE

Typically one assumes the null hypothesis to be true and then one of the following conclusions are drawn.

1. Reject H0 Equivalent to saying that Ha is correct or true

2. Fail to reject H0 Equivalent to saying that we have failed to show a statistically significant deviation from the claim of the null hypothesis


AN ANALOGY

The Statistical Hypothesis Testing process can be compared very closely with a judicial trial.

Assume a defendant is innocent (H0)

Present evidence to show guilt

Try to prove guilt beyond a reasonable doubt(Ha)


AN ANALOGY

Two Hypotheses are then created.

H0: Innocent

Ha: Not Innocent (Guilt)


Examples of Hypotheses

You would like to determine if the diameters of the ball bearings you produce have a mean of 6.5 cm.

H0: =6.5

Ha: 6.5

(Two-sided alternative)


The students entering into the math program used to have a mean SAT quantitative score of 525. Are the current students poorer (as measured by the SAT quantitative score)?

H0: = 525

(Really:

Ha: < 525

(One-sided alternative)



Do the “16 ounce” cans of peaches canned and sold by DelMonte meet the claim on the label (on the average)?

Notice, the real concern would be selling the consumer less than 16 ounces of peaches.

H0: = 16

(Really:

Ha: < 16



Is the proportion of defective parts produced by a manufacturing process more than 5%?

H0: = 0.05

(Really,

Ha: > 0.05



Do two brands of light bulb

have the same mean

lifetime?

H0: Brand A = Brand B

Ha: Brand A Brand B



Do parts produced by two

different milling machines

have the same variability

in diameters.

or equivalently


0 1 2

a 1 2

H :

H :

2 2

0 1 2

2 2

a 1 2

H :

H :


Comments on Hypothesis Form

The null hypothesis must contain the equal sign.

This is absolutely necessary because the test requires the null hypothesis to be assumed to be true and the value attached to the equal sign is then the value assumed to be true.

The alternate hypothesis should be what you are really attempting to show to be true.

This is not always possible.


Hypothesis FormThe form of the null hypothesis is

H0: population characteristic = hypothesized value

Where the hypothesized value is a specific number determined by the problem context.

The alternative (or alternate) hypothesis will have one of the following three forms:

Ha: population characteristic > hypothesized value

Ha: population characteristic < hypothesized value

Ha: population characteristic hypothesized value


Caution

When you set up a hypothesis test, the result is eitherStrong support for the alternate hypothesis

(if the null hypothesis is rejected)There is not sufficient evidence to refute

the claim of the null hypothesis (you are stuck with it, but there is only a lack of strong evidence against the null hypothesis.


Error

Null Hypothesis

Decision True False

Accept H0

Reject H0

No Error

No ErrorType I Error

Type II Error


Error AnalogyConsider a medical test where the hypotheses are equivalent to

H0: the patient has a specific disease

Ha: the patient doesn’t have the disease

Then,Type I error is equivalent to a false negative(I.e., Saying the patient does not have the disease when in

fact, he does.)Type II error is equivalent to a false positive(I.e., Saying the patient has the disease when, in fact, he

does not.)


More on Error

The probability of a type I error is denoted by and is called the level of significance of the test. Thus, a test with = 0.01 is said to have a level of significance of 0.01 or to be a level 0.01 test.

The probability of a type II error is denoted by .


Relationships Between and Generally, with everything else held

constant, decreasing one type of error causes the other to increase.

The only way to decrease both types of error simultaneously is to increase the sample size.

No matter what decision is reached, there is always the risk of one of these errors.


Comment of ProcessLook at the consequences of type I and type II errors and then identify the largest that is tolerable for the problem.

Employ a test procedure that uses this maximum acceptable value of (rather than anything smaller) as the level of significance (because using a smaller increases ).


Test Statistic

A test statistic is the function of sample data on which a conclusion to reject or fail to reject H0 is based.


P-value

The P-value (also called the observed significance level) is a measure of inconsistency between the hypothesized value for a population characteristic and the observed sample.

The P-value is the probability, assuming that H0 is true, of obtaining a test statistic value at least as inconsistent with H0 as what actually resulted.


Decision Criteria

A decision as to whether H0 should be rejected results from comparing the P-value to the chosen a:

H0 should be rejected if P-value

H0 should not be rejected if P-value >


Large Sample Hypothesis Test for a Single Proportion

In terms of a standard normal random variable z, the approximate P-value for this test is

0

0 0

pz

(1 )

n

To test the hypothesis H0: =0, compute the z statistic

where 0 is the hypothesized proportion.

0

0 0

pz

(1 )

n


where 0 is the hypothesized proportion.


0 stands for the hypothesized value of the population proportion from the null hypothesis.

Hypothesis Test Large Sample Test of Population Proportion

0a 0

0 0

pP-value P z if H :

(1 )n




0a 0

0 0

pP-value P z if H :

(1 )n




0a 0

0 0

pP-value 2P z if H :

(1 )n


An insurance company states that the proportion of its claims that are settled within 30 days is 0.9. A consumer group thinks that the company drags its feet and takes longer to settle claims. To check these hypotheses, a simple random sample of 200 of the company’s claims was obtained and it was found that 160 of the claims were settled within 30 days.

Hypothesis Test Example ( Large-Sample Test for a Population Proportion)


Hypothesis Test Example 2 ( Single Proportion) continued

= proportion of the company’s claims that are settled within 30 days

H0: = 0.9Ha: 0.9

P(p 0.8 if 0.9) P(z 4.71) 0

p 0.9 0.8 0.9z 4.71

(0.9)(0.1) (0.9)(0.1)200 200

160p= 0.8

200The sample proportion is

160p= 0.8

200The sample proportion is



The probability of getting a result as strongly or more strongly in favor of the consumer group's claim (the alternate hypothesis Ha if the company’s claim (H0) was true is essentially 0.

Clearly, this gives strong evidence in support of the alternate hypothesis (against the null hypothesis).



We would say that we have strong support for the claim that the proportion of the insurance company’s claims that are settled within 30 days is less than 0.9.

Some people would state that we have shown that the true proportion of the insurance company’s claims that are settled within 30 days is statistically significantly less than 0.9.


A county judge has agreed that he will give up his county judgeship and run for a state judgeship unless there is evidence at the 0.10 level that more then 25% of his party is in opposition. A SRS of 800 party members included 217 who opposed him. Please advise this judge.

Hypothesis Test Example( Single Proportion)


Hypothesis Test Example( Single Proportion) continued

= proportion of his party that is in opposition

H0: = 0.25Ha: > 0.25= 0.10Note: hypothesized value = 0 = 0.25

217n 800, p 0.27125

800

0.27125 0.25z 1.39

0.25(0.75)800


At a level of significance of 0.10, there is sufficient evidence to support the claim that the true percentage of the party members that oppose him is more than 25%.

Under these circumstances, I would advise him not to run.

Hypothesis Test Example( Single Proportion) continued

P-value P(z 1.39 if actually equals 0.25)

1 0.9177 0.0823


Steps in a Hypothesis-Testing Analysis

1. Describe (determine) the population characteristic about which hypotheses are to be tested.

2. State the null hypothesis H0.

3. State the alternate hypothesis Ha.

4. Select the significance level for the test.

5. Display the test statistic to be used, with substitution of the hypothesized value identified in step 2 but without any computation at this point.


Steps in a Hypothesis-Testing Analysis

6. Check to make sure that any assumptions required for the test are reasonable.

7. Compute all quantities appearing in the test statistic and then the value of the test statistic itself.

8. Determine the P-value associated with the observed value of the test statistic

9. State the conclusion in the context of the problem, including the level of significance.


Hypothesis Test (Large samples)Single Sample Test of Population Mean

In terms of a standard normal random variable z, the approximate P-value for this test is

0xz

n


where 0 is the hypothesized mean.

0xz

n




0 stands for the hypothesized value of the population mean from the null hypothesis.

Hypothesis Test Single Sample Test of Population Mean

0a 0

xP-value P z if H :

n




0a 0

xP-value P z if H :

n




0a 0

xP-value 2P z if H :

n


Reality CheckIt is not likely that one would know but not know so calculating a z-value using the formula

would not be very realistic. For large values of n (>30) it is generally acceptable to use s to estimate , however, it is much more common to apply the t-distribution.

0xz

n

It is not likely that one would know but not know so calculating a z-value using the formula

would not be very realistic. For large values of n (>30) it is generally acceptable to use s to estimate , however, it is much more common to apply the t-distribution.

0xz

n


Hypothesis Test ( unknown) Single Sample Test of Population Mean

In terms of the t random variable with degrees of freedom df = n-1, the approximate P-value for this test is

To test the hypothesis H0: =0, when we may assume that the underlying distribution is approximately (or actually normal)compute the t statistic


0xt

sn

To test the hypothesis H0: =0, when we may assume that the underlying distribution is approximately (or actually normal)compute the t statistic


0xt

sn


The t statistic can be used for all sample sizes, however, the smaller the sample, the more important the assumption that the underlying distribution is normal.

Typically, when n >15 the underlying distribution need only be centrally weighted and may be somewhat skewed.

Hypothesis Test ( unknown) Single Sample Test of Population Mean


Tail areas for t curves

t df 1 2 3 4 5 6 7 8 9 10 11 12

0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.5000.1 0.468 0.465 0.463 0.463 0.462 0.462 0.462 0.461 0.461 0.461 0.461 0.4610.2 0.437 0.430 0.427 0.426 0.425 0.424 0.424 0.423 0.423 0.423 0.423 0.4220.3 0.407 0.396 0.392 0.390 0.388 0.387 0.386 0.386 0.385 0.385 0.385 0.3850.4 0.379 0.364 0.358 0.355 0.353 0.352 0.351 0.350 0.349 0.349 0.348 0.3480.5 0.352 0.333 0.326 0.322 0.319 0.317 0.316 0.315 0.315 0.314 0.313 0.313

0.6 0.328 0.305 0.295 0.290 0.287 0.285 0.284 0.283 0.282 0.281 0.280 0.2800.7 0.306 0.278 0.267 0.261 0.258 0.255 0.253 0.252 0.251 0.250 0.249 0.2490.8 0.285 0.254 0.241 0.234 0.230 0.227 0.225 0.223 0.222 0.221 0.220 0.2200.9 0.267 0.232 0.217 0.210 0.205 0.201 0.199 0.197 0.196 0.195 0.194 0.193

1.0 0.250 0.211 0.196 0.187 0.182 0.178 0.175 0.173 0.172 0.170 0.169 0.1691.1 0.235 0.193 0.176 0.167 0.161 0.157 0.154 0.152 0.150 0.149 0.147 0.1461.2 0.221 0.177 0.158 0.148 0.142 0.138 0.135 0.132 0.130 0.129 0.128 0.1271.3 0.209 0.162 0.142 0.132 0.125 0.121 0.117 0.115 0.113 0.111 0.110 0.1091.4 0.197 0.148 0.128 0.117 0.110 0.106 0.102 0.100 0.098 0.096 0.095 0.0931.5 0.187 0.136 0.115 0.104 0.097 0.092 0.089 0.086 0.084 0.082 0.081 0.080

1.6 0.178 0.125 0.104 0.092 0.085 0.080 0.077 0.074 0.072 0.070 0.069 0.0681.7 0.169 0.116 0.094 0.082 0.075 0.070 0.066 0.064 0.062 0.060 0.059 0.0571.8 0.161 0.107 0.085 0.073 0.066 0.061 0.057 0.055 0.053 0.051 0.050 0.0491.9 0.154 0.099 0.077 0.065 0.058 0.053 0.050 0.047 0.045 0.043 0.042 0.0412.0 0.148 0.092 0.070 0.058 0.051 0.046 0.043 0.040 0.038 0.037 0.035 0.034

2.1 0.141 0.085 0.063 0.052 0.045 0.040 0.037 0.034 0.033 0.031 0.030 0.0292.2 0.136 0.079 0.058 0.046 0.040 0.035 0.032 0.029 0.028 0.026 0.025 0.0242.3 0.131 0.074 0.052 0.041 0.035 0.031 0.027 0.025 0.023 0.022 0.021 0.0202.4 0.126 0.069 0.048 0.037 0.031 0.027 0.024 0.022 0.020 0.019 0.018 0.0172.5 0.121 0.065 0.044 0.033 0.027 0.023 0.020 0.018 0.017 0.016 0.015 0.014

2.6 0.117 0.061 0.040 0.030 0.024 0.020 0.018 0.016 0.014 0.013 0.012 0.0122.7 0.113 0.057 0.037 0.027 0.021 0.018 0.015 0.014 0.012 0.011 0.010 0.0102.8 0.109 0.054 0.034 0.024 0.019 0.016 0.013 0.012 0.010 0.009 0.009 0.0082.9 0.106 0.051 0.031 0.022 0.017 0.014 0.011 0.010 0.009 0.008 0.007 0.0073.0 0.102 0.048 0.029 0.020 0.015 0.012 0.010 0.009 0.007 0.007 0.006 0.006

3.1 0.099 0.045 0.027 0.018 0.013 0.011 0.009 0.007 0.006 0.006 0.005 0.0053.2 0.096 0.043 0.025 0.016 0.012 0.009 0.008 0.006 0.005 0.005 0.004 0.0043.3 0.094 0.040 0.023 0.015 0.011 0.008 0.007 0.005 0.005 0.004 0.004 0.0033.4 0.091 0.038 0.021 0.014 0.010 0.007 0.006 0.005 0.004 0.003 0.003 0.0033.5 0.089 0.036 0.020 0.012 0.009 0.006 0.005 0.004 0.003 0.003 0.002 0.002

3.6 0.086 0.035 0.018 0.011 0.008 0.006 0.004 0.003 0.003 0.002 0.002 0.0023.7 0.084 0.033 0.017 0.010 0.007 0.005 0.004 0.003 0.002 0.002 0.002 0.0023.8 0.082 0.031 0.016 0.010 0.006 0.004 0.003 0.003 0.002 0.002 0.001 0.0013.9 0.080 0.030 0.015 0.009 0.006 0.004 0.003 0.002 0.002 0.001 0.001 0.0014.0 0.078 0.029 0.014 0.008 0.005 0.004 0.003 0.002 0.002 0.001 0.001 0.001



t df 25 26 27 28 29 30 35 40 60 120 (=z)

0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.5000.1 0.461 0.461 0.461 0.461 0.461 0.461 0.460 0.460 0.460 0.460 0.4600.2 0.422 0.422 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.421 0.4210.3 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.383 0.382 0.3820.4 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.346 0.345 0.345 0.3450.5 0.311 0.311 0.311 0.310 0.310 0.310 0.310 0.310 0.309 0.309 0.309

0.6 0.277 0.277 0.277 0.277 0.277 0.277 0.276 0.276 0.275 0.275 0.2740.7 0.245 0.245 0.245 0.245 0.245 0.245 0.244 0.244 0.243 0.243 0.2420.8 0.216 0.215 0.215 0.215 0.215 0.215 0.215 0.214 0.213 0.213 0.2120.9 0.188 0.188 0.188 0.188 0.188 0.188 0.187 0.187 0.186 0.185 0.184

1.0 0.163 0.163 0.163 0.163 0.163 0.163 0.162 0.162 0.161 0.160 0.1591.1 0.141 0.141 0.141 0.140 0.140 0.140 0.139 0.139 0.138 0.137 0.1361.2 0.121 0.120 0.120 0.120 0.120 0.120 0.119 0.119 0.117 0.116 0.1151.3 0.103 0.103 0.102 0.102 0.102 0.102 0.101 0.101 0.099 0.098 0.0971.4 0.087 0.087 0.086 0.086 0.086 0.086 0.085 0.085 0.083 0.082 0.0811.5 0.073 0.073 0.073 0.072 0.072 0.072 0.071 0.071 0.069 0.068 0.067

1.6 0.061 0.061 0.061 0.060 0.060 0.060 0.059 0.059 0.057 0.056 0.0551.7 0.051 0.051 0.050 0.050 0.050 0.050 0.049 0.048 0.047 0.046 0.0451.8 0.042 0.042 0.042 0.041 0.041 0.041 0.040 0.040 0.038 0.037 0.0361.9 0.035 0.034 0.034 0.034 0.034 0.034 0.033 0.032 0.031 0.030 0.0292.0 0.028 0.028 0.028 0.028 0.027 0.027 0.027 0.026 0.025 0.024 0.023

2.1 0.023 0.023 0.023 0.022 0.022 0.022 0.022 0.021 0.020 0.019 0.0182.2 0.019 0.018 0.018 0.018 0.018 0.018 0.017 0.017 0.016 0.015 0.0142.3 0.015 0.015 0.015 0.015 0.014 0.014 0.014 0.013 0.012 0.012 0.0112.4 0.012 0.012 0.012 0.012 0.012 0.011 0.011 0.011 0.010 0.009 0.0082.5 0.010 0.010 0.009 0.009 0.009 0.009 0.009 0.008 0.008 0.007 0.006

2.6 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.006 0.005 0.0052.7 0.006 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.004 0.004 0.0032.8 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.003 0.003 0.0032.9 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.002 0.0023.0 0.003 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.001

3.1 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.0013.2 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.0013.3 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.0003.4 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.0003.5 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000

3.6 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.0003.7 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0003.8 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0003.9 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.0004.0 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000




0a 0

xP-value P t if H :

sn




0a 0

xP-value P t if H :

sn




0a 0

xP-value 2P t if H :

sn


An manufacturer of a special bolt requires that this type of bolt have a mean shearing strength in excess of 110 lb. To determine if the manufacturer’s bolts meet the required standards a sample of 25 bolts was obtained and tested. The sample mean was 112.7 lb and the sample standard deviation was 9.62 lb. Use this information to perform an appropriate hypothesis test with a significance level of 0.05.

Example of Hypothesis TestSingle Sample Test of Population Mean continued


= the mean shearing strength of this specific type of bolt

he hypotheses to be tested are H0: =110 lb HA: 110 lb

The significance level to be used for the test is = 0.05.


x hypothesized meant

sn

The test statistic is x hypothesized meant

sn

The test statistic is



P-value P(x 112.7 if is actually 110)

112.7 110P t

9.6225

P(t 1.4) 0.087

x 112.7, s 9.62, n 25, df 24


Because P-value = 0.087 > 0.05 = we fail to reject H0.

At a level of significance of 0.05, there is insufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 110 lbs.

Example of Hypothesis TestSingle Sample Test of Population Mean conclusion



t df 13 14 15 16 17 18 19 20 21 22 23 24

0.0 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.5000.1 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.461 0.4610.2 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.422 0.4220.3 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.384 0.383 0.383 0.3830.4 0.348 0.348 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.347 0.346 0.3460.5 0.313 0.312 0.312 0.312 0.312 0.312 0.311 0.311 0.311 0.311 0.311 0.311

0.6 0.279 0.279 0.279 0.278 0.278 0.278 0.278 0.278 0.277 0.277 0.277 0.2770.7 0.248 0.248 0.247 0.247 0.247 0.246 0.246 0.246 0.246 0.246 0.245 0.2450.8 0.219 0.219 0.218 0.218 0.217 0.217 0.217 0.217 0.216 0.216 0.216 0.2160.9 0.192 0.192 0.191 0.191 0.190 0.190 0.190 0.189 0.189 0.189 0.189 0.189

1.0 0.168 0.167 0.167 0.166 0.166 0.165 0.165 0.165 0.164 0.164 0.164 0.1641.1 0.146 0.145 0.144 0.144 0.143 0.143 0.143 0.142 0.142 0.142 0.141 0.1411.2 0.126 0.125 0.124 0.124 0.123 0.123 0.122 0.122 0.122 0.121 0.121 0.1211.3 0.108 0.107 0.107 0.106 0.105 0.105 0.105 0.104 0.104 0.104 0.103 0.1031.4 0.092 0.092 0.091 0.090 0.090 0.089 0.089 0.088 0.088 0.088 0.087 0.0871.5 0.079 0.078 0.077 0.077 0.076 0.075 0.075 0.075 0.074 0.074 0.074 0.073

1.6 0.067 0.066 0.065 0.065 0.064 0.064 0.063 0.063 0.062 0.062 0.062 0.0611.7 0.056 0.056 0.055 0.054 0.054 0.053 0.053 0.052 0.052 0.052 0.051 0.0511.8 0.048 0.047 0.046 0.045 0.045 0.044 0.044 0.043 0.043 0.043 0.042 0.0421.9 0.040 0.039 0.038 0.038 0.037 0.037 0.036 0.036 0.036 0.035 0.035 0.0352.0 0.033 0.033 0.032 0.031 0.031 0.030 0.030 0.030 0.029 0.029 0.029 0.028

2.1 0.028 0.027 0.027 0.026 0.025 0.025 0.025 0.024 0.024 0.024 0.023 0.0232.2 0.023 0.023 0.022 0.021 0.021 0.021 0.020 0.020 0.020 0.019 0.019 0.0192.3 0.019 0.019 0.018 0.018 0.017 0.017 0.016 0.016 0.016 0.016 0.015 0.0152.4 0.016 0.015 0.015 0.014 0.014 0.014 0.013 0.013 0.013 0.013 0.012 0.0122.5 0.013 0.013 0.012 0.012 0.011 0.011 0.011 0.011 0.010 0.010 0.010 0.010

2.6 0.011 0.010 0.010 0.010 0.009 0.009 0.009 0.009 0.008 0.008 0.008 0.0082.7 0.009 0.009 0.008 0.008 0.008 0.007 0.007 0.007 0.007 0.007 0.006 0.0062.8 0.008 0.007 0.007 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.0052.9 0.006 0.006 0.005 0.005 0.005 0.005 0.005 0.004 0.004 0.004 0.004 0.0043.0 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.003 0.003 0.003 0.003

3.1 0.004 0.004 0.004 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.0023.2 0.003 0.003 0.003 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.0023.3 0.003 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.0023.4 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.0013.5 0.002 0.002 0.002 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001

3.6 0.002 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.0013.7 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.0013.8 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.0003.9 0.001 0.001 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.0004.0 0.001 0.001 0.001 0.001 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

t=1.4

n=25

df=24

Tail area = 0.087


Revisit the problem with = 0.10

What would happen if the significance level of the test was 0.10 instead of 0.05?


Now P-value= 0.087 < 0.10=and we reject H0 at the 0.10 level of significance and conclude

At the 0.10 level of significance there is sufficient evidence to conclude that the mean shearing strength of this brand of bolt exceeds 120 lbs.

Revisit the problem with = 0.10


Comments continued

Many people are bothered by the fact that different choices of lead to different conclusions.

This is nature of a process where you control the probability of being wrong when you select the level of significance. This reflects your willingness to accept a certain level of type I error.


Another ExampleA jeweler is planning on manufacturing gold charms. His design calls for a particular piece to contain 0.08 ounces of gold. The jeweler would like to know if the pieces that he makes contain (on the average) 0.08 ounces of gold. To test to see if the pieces contain 0.08 ounces of gold, he made a sample of 16 of these particular pieces and obtained the following data. 0.0773 0.0779 0.0756 0.0792 0.0777 0.0713 0.0818 0.0802 0.0802 0.0785 0.0764 0.0806 0.0786 0.0776 0.0793 0.0755Use a level of significance of 0.01 to perform an appropriate hypothesis test.


Another Example1. The population characteristic being

studied is = true mean gold content for this particular type of charm.

2. Null hypothesis: H0: = 0.08 oz

3. Alternate hypothesis: Ha: 0.08 oz

4. Significance level: = 0.01x hypothesized mean x 0.08

ts s

n n

5. Test statistic:

x hypothesized mean x 0.08t

s sn n

5. Test statistic:


Another Example6. Minitab was used to create a normal

plot along with a graphical display of the descriptive statistics for the sample data.

P-Value: 0.396A-Squared: 0.363

Anderson-Darling Normality Test

N: 16StDev: 0.0025143Average: 0.0779813

0.0820.0770.072

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

bab

ility

Gold

Normal Probability Plot


Another ExampleWe can see that with the exception of one outlier, the data is reasonably symmetric and mound shaped in shape, indicating that the assumption that the population of amounts of gold for this particular charm can reasonably be expected to be normally distributed.

0.0820.0800.0780.0760.0740.072

95% Confidence Interval for Mu

0.07950.07850.07750.0765

95% Confidence Interval for Median

Variable: Gold

7.71E-02

1.86E-03

7.66E-02

Maximum3rd QuartileMedian1st QuartileMinimum

NKurtosisSkewnessVarianceStDevMean

P-Value:A-Squared:

7.95E-02

3.89E-03

7.93E-02

8.18E-028.00E-027.82E-027.66E-027.13E-02

162.23191

-1.109226.32E-062.51E-037.80E-02

0.3960.363

95% Confidence Interval for Median

95% Confidence Interval for Sigma

95% Confidence Interval for Mu

Anderson-Darling Normality Test

Descriptive Statistics


Another Example

8. P-value: This is a two tailed test. Looking up in the table of tail areas for t curves, t = 3.2 with df = 15 we see the table entry is 0.003 so

P-Value = 2(0.003) = 0.006

n 16, x 0.077981, s 0.0025143

0.077981 0.08t 3.2

0.002514316

7. Computations:

n 16, x 0.077981, s 0.0025143

0.077981 0.08t 3.2

0.002514316

7. Computations:


Another Example9. Conclusion:

Since P-value = 0.006 0.01 = , we reject H0 at the 0.01 level of significance.At the 0.01 level of significance there is convincing evidence that the true mean gold content of this type of charm is not 0.08 ounces. [Actually when rejecting a null hypothesis for alternative, a one tailed claim is supported. In this case, at the 0.01 level of significance, there is convincing evidence that the true mean gold content of this type of charm is less than 0.08 ounces.


Power and Probability of Type II Error

The power of a test is the probability of rejecting the null hypothesis.

When H0 is false, the power is the probability that the null hypothesis is rejected. Specifically, power = 1 – .


Effects of Various Factors on Power

1. The larger the size of the discrepancy between the hypothesized value and the true value of the population characteristic, the higher the power.

2. The larger the significance level, , the higher the power of the test.

3. The larger the sample size, the higher the power of the test.


Some Comments

Calculating (hence power) depends on knowing the true value of the population characteristic being tested. Since the true value is not known, generally, one calculates for a number of possible “true” values of the characteristic under study and then sketches a power curve.


Example (based on z-curve)

Consider the earlier example where we tested H0: = 110 vs. Ha: u > 110 and furthermore, suppose the true standard deviation of the bolts was actually 10 lbs.



Power CurvesDifferent 's

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

118 119 120 121 122 123 124 125 126 127 128

True Value of

Po

we

r (1

-

)

H0: = 120

Ha: > 120 = 10



Power CurvesDifferent n's

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

118 119 120 121 122 123 124 125 126 127 128

True Value of

Po

we

r (1

-

)

n = 45

n = 90

n = 180

n = 360

H0: = 120

Ha: > 120 = 10

Documents

Chapter 10 Hypothesis Tests Using a Single Sample