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Question 1 (10 points)
Part (a) (4 points)
Chicken Delight claims that the mean wait time for an order is 5 minutes or less. A sample
of 50 orders revealed a mean wait time of 5.2 minutes. The standard deviation is 1 minute.
Using a level of significance of 0.05, carry out a hypothesis test for Chicken Delight’s
claim.
Solution
First, we state null and alternative hypothesis
Null hypothesis: The mean wait time for an order is 5 minutes or less
Alternative hypothesis: The mean wait time for an order is more than 5 minutes
Mathematical presentation of hypothesis
: 5
: 5
Since n>30 , we use z test
Z =

󰇡
󰇢

=

󰇡

󰇢
= 1.414
At 5% level of significance, Z critical is calculated
Z critical (

󰇜 = 1.645
The value of

(1.414) is less than Z critical value hence we fail to reject null hypothesis.
Hethe the nce that we do not have enough evidence to conclude that the mean wait time for an
order is more than 5 minutes and therefore delights claim is very true.
Part (b) (3 points)
Chicken Delight claims that the mean wait time for an order is 5 minutes or less. A sample
of 200 orders revealed a mean wait time of 5.2 minutes. The standard deviation is 1 minute.
Using a level of significance of 0.05, carry out a hypothesis test for Chicken Delight’s claim.
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Solution
First, we state null and alternative hypothesis
Null hypothesis: The mean wait time for an order is 5 minutes or less
Alternative hypothesis: The mean wait time for an order is more than 5 minutes
Mathematical presentation of hypothesis
: 5
: 5
Since n>30 , we use z test
Z =

󰇡
󰇢

=

󰇡

󰇢
= 2.828
At 5% level of significance, Z critical is calculated
Z critical (

󰇜 = 1.645
The value of

󰇛󰇜 is greater than Z critical (1.645) value hence we reject null
hypothesis. Ththe the e results are statistically significant at 0.05 level of significance and we say
that there is enough evidence to conclude that the mean wait time for an order is more than 5
minutes and therefore delights claim is false.
Part (c) (3 points)
Compare your results from Parts (a) and (b) and provide an explanation for what you find.
The results in part (a) showed that the results are statistically insignificant while in part (b) the
results are statistically significant. In part (a), we failed to reject the null hypothesis while in part
(b) we rejected the null hypothesis.
Explanation
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Increasing sample size increases the accuracy of test results. After increasing sample size from
50 to 200, the test results became significant at 5% level of significance. Therefore, we can
conclude that large sample size improves the accuracy of test results.
Question 2 (10 points)
Part (a) (5 points)
The following information is given:
H
0
is that the mean speed of cars driving on the Garden State Parkway is 60 miles
per hour or less.
The sample consists of 359 cars driving on the Garden State Parkway. Their mean
speed is 63 miles per hour.
α = 0.1
If it is possible to determine the critical value(s) based on this information, determine
it/them. If it is not possible, explain why it is not possible.
It is possible to calculate the critical value.
We are testing the hypothesis
: 60
: 60
This is a right-tailed test.
The sample size = 359 which is greater than 30 hence we use z test
0.1

= 1.28 (from standard normal table)
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Part (b) (5 points)
Part (b) is independent of Part (a), i.e., do not use information from Part (a) to answer Part (b).
The following information is given:
The null hypothesis is that the population mean of middle school teachers’ salaries is
\$49,500.
The significance level is 0.05.
You know a researcher has collected a sample of 136 middle school teachers’
salaries, but you do not know what the mean salary in the sample is.
If it is possible to determine the critical value(s) based on this information,
determine it/them. If it is not possible, explain why it is not possible.
It is possible to calculate the critical value.
We are testing the hypothesis
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: =49500
: 49500
This is a two-tailed test.
The sample size = 136 which is greater than 30 hence we use z test
0.05

=

= 1.96 (from the standard normal table)
Question 3 (5 points)
Suppose that you want to test the claim that the mean thickness of an iPad screen is 2.1
millimeters. You collect a sample of 80 iPads to carry out a hypothesis test, using a
significance level (alpha) of 0.1. Your decision is to not reject the null hypothesis.
Part (a) (2.5 points)
Using the information above, what is the largest possible p-value for this scenario?
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The null hypothesis is rejected when P(

>

)
The largest possible p-value is given by P󰇧

󰇡

󰇢
󰇨 where S is the standard deviation
Part (b) (2.5 points)
Using the information above, what is the smallest possible p-value for this scenario?
The null hypothesis is rejected when P(

<

)
The smallest possible p-value is given by P󰇧

󰇡

󰇢
󰇨 where S is the standard deviation

Unformatted Attachment Preview

Question 1 (10 points) Part (a) (4 points) Chicken Delight claims that the mean wait time for an order is 5 minutes or less. A sample of 50 orders revealed a mean wait time of 5.2 minutes. The standard deviation is 1 minute. Using a level of significance of 0.05, carry out a hypothesis test for Chicken Delight’s claim. Solution First, we state null and alternative hypothesis Null hypothesis: The mean wait time for an order is 5 minutes or less Alternative hypothesis: The mean wait time for an order is more than 5 minutes Mathematical presentation of hypothesis 𝐻0 : 𝜇 ≤ 5 𝐻1 : 𝜇 > 5 Since n>30 , we use z test Z= 𝑥̅ −𝜇 ( 𝜎 ) √𝑛 𝑍𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 = 5.2−5 ( 1 ) √50 = 1.414 At 5% level of significance, Z critical is calculated Z critical (𝑧0.05 ) = 1.645 The value of 𝑍𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 (1.414) is less than Z critical value hence we fail to reject null hypothesis. Hethe the nce that we do not have enough evidence to conclude that the mean wait time for an order is more than 5 minutes and therefore delights claim is very true. Part (b) (3 points) Chicken Delight claims that the mean wait time for an order is 5 minutes or less. A sample of 200 orders revealed a mean wait time of 5.2 minutes. The standard deviation is 1 minute. Using a level of significance of 0.05, carry out a hypothesis test for Chicken Delight’s claim. Solution First, we state null and alternative hypothesis Null hypothesis: The mean wait time for an order is 5 minutes or less Alternative hypothesis: The mean wait time for an order is more than 5 minutes Mathematical presentation of hypothesis 𝐻0 : 𝜇 ≤ 5 𝐻1 : 𝜇 > 5 Since n>30 , we use z test Z= 𝑥̅ −𝜇 ( 𝜎 ) √𝑛 𝑍𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 = 5.2−5 ( 1 ) √200 = 2.828 At 5% level of significance, Z critical is calculated Z critical (𝑧0.05 ) = 1.645 The value of 𝑍𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 (2.828) is greater than Z critical (1.645) value hence we reject null hypothesis. Ththe the e results are statistically significant at 0.05 level of significance and we say that there is enough evidence to conclude that the mean wait time for an order is more than 5 minutes and therefore delights claim is false. Part (c) (3 points) Compare your results from Parts (a) and (b) and provide an explanation for what you find. The results in part (a) showed that the results are statistically insignificant while in part (b) the results are statistically significant. In part (a), we failed to reject the null hypothesis while in part (b) we rejected the null hypothesis. Explanation Increasing sample size increases the accuracy of test results. After increasing sample size from 50 to 200, the test results became significant at 5% level of significance. Therefore, we can conclude that large sample size improves the accuracy of test results. Question 2 (10 points) Part (a) (5 points) The following information is given: • H0 is that the mean speed of cars driving on the Garden State Parkway is 60 miles per hour or less. • The sample consists of 359 cars driving on the Garden State Parkway. Their mean speed is 63 miles per hour. α = 0.1 If it is possible to determine the critical value(s) based on this information, determine it/them. If it is not possible, explain why it is not possible. It is possible to calculate the critical value. We are testing the hypothesis 𝐻0 : 𝜇 ≤ 60 𝐻1 : 𝜇 > 60 This is a right-tailed test. The sample size = 359 which is greater than 30 hence we use z test 𝛼 =0.1 𝑍0.1 = 1.28 (from standard normal table) Part (b) (5 points) Part (b) is independent of Part (a), i.e., do not use information from Part (a) to answer Part (b). The following information is given: • The null hypothesis is that the population mean of middle school teachers’ salaries is \$49,500. • The significance level • You is 0.05. know a researcher has collected a sample of 136 middle school teachers’ salaries, but you do not know what the mean salary in the sample is. If it is possible to determine the critical value(s) based on this information, determine it/them. If it is not possible, explain why it is not possible. It is possible to calculate the critical value. We are testing the hypothesis 𝐻0 : 𝜇=49500 𝐻1 : 𝜇 ≠ 49500 This is a two-tailed test. The sample size = 136 which is greater than 30 hence we use z test 𝛼 =0.05 𝑍0.05 = 𝑍0.025 = 1.96 (from the standard normal table) 2 Question 3 (5 points) Suppose that you want to test the claim that the mean thickness of an iPad screen is 2.1 millimeters. You collect a sample of 80 iPads to carry out a hypothesis test, using a significance level (alpha) of 0.1. Your decision is to not reject the null hypothesis. Part (a) (2.5 points) Using the information above, what is the largest possible p-value for this scenario? The null hypothesis is rejected when P(𝑍𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 > 𝑍𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 ) The largest possible p-value is given by P(Z > Part (b) −2.1 ( 𝑆 ) √80 ) where S is the standard deviation (2.5 points) Using the information above, what is the smallest possible p-value for this scenario? The null hypothesis is rejected when P(𝑍𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 < 𝑍𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 ) The smallest possible p-value is given by P(Z < − −2.1 ( 𝑆 ) √80 ) where S is the standard deviation Name: Description: ...
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