Homework 8 Bus Stats
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Feb 20, 2024
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MAT-1430 Business Statistics
Homework 8
1.
(4 points) A researcher believes that more than 70 percent of people would prefer watching a movie on a streaming service at home than in a movie theater. Suppose a random sample of 1,000 people indicates that 73% prefer watching movies in their home. Perform the appropriate significance test to comment on the researcher’s claim. Use a significance level of 0.025.
Independence: Each individual should be independent of one another.
Randomization: States random sample
10% condition: 1,000 people is less than 10% of all people who watch movies.
Large enough: 1,000>30 so the distribution is nearly normal. Ho: p <= .7
Ha: p > .7
N = 1000
P(hat) = .73
A = .025
Z* = 1.962
Z = (.73-.7)/sqrt((.7*.3)/1000) = 2.07
Reject the null
Comparing 2.07 to the cutoff of 1.962 there is enough evidence to suggest that more than 70% of people would prefer to watch a movie on a streaming service at home than in a movie theatre. 2.
(4 points) Suppose the IRS posed as taxpayers to test certain taxpayer assistance employees. The IRA reported that, upon asking the assistant a tax related question, they gave the correct response 69% of the time. Under suspicion that this number may actually be higher or lower, an independent commission randomly sampled 200 tax returns completed with the assistance of IRS employees. The number of accurate returns in this sample was 125. Perform a significance test to determine if evidence exists that the true accuracy rate is higher or lower than the 69% reported. Use a significance level of 0.01 Independence: Each taxpayer should be independent of one another.
Randomization: States random sample
10% condition: 200 people is less than 10% of all people who pay taxes
Large enough: 200>30 so the distribution is nearly normal. Ho: p = .69
Ha: p =/ .69 (claim)
N = 200
P(hat) = 125/200 = .625
A = .01
Z = (.625-.69)/sqrt((.69*.31)/200) = -1.9875 = .0233
Fail to reject
With a p value of .0233 at the .01/2 significance level there is not enough evidence to suggest that the true proportion accuracy rate is higher or lower than the 69% rate reported. 3.
(4 points) A comparison between two tax preparation software packages is needed to determine if there’s a difference in time when filing taxes. 55 random people were assigned the task of filing their taxes, once with Turbo-Tax and once with a
competitor. (Half were assigned to use Turbo-Tax first, the other half the competitor.) The competing software package’s
average completion time was 6 minutes longer and the differences had a standard deviation of 12 minutes. What conclusion can be made regarding the true mean difference in time to file taxes? Use a significance level of 0.05.
Independence: Each tax filer should be independent of one another.
Randomization: States random sample
10% condition: 55 people is less than 10% of all people who file taxes
Large enough: 55>30 so the distribution is nearly normal. Ho: Mu = 0
Ha: Mu =/ 0
X = 6
SD = 12
N = 55
A = .05
T* = (6-0)/(12/sqrt(55)) = 3.71
T = 2.009
Reject the null
Considering 3.71 at a cutoff of 2.009 there is enough evidence to suggest that there is a difference in time to file taxes between the two companies. 4.
(3 points) Construct and interpret a 95% confidence interval for the previous example.
6 +- 2.009(12/sqrt(55)) = (2.7493, 9.2507)
We are 95% confident that the true mean difference in the time taken in filing taxes between the two companies is between
2.7493 and 9.2507 minutes. 5.
(6 points) A credit card company operates two call centers. In order to determine if there’s a difference between the average length of a customer service call for the two centers, the following random sample results were recorded:
Center A
Center B
n = 36 calls
n = 32 calls
x
= 5.2 mins
x
= 6.2 mins
s = 1.2 mins
s = 2.1 mins
a.
Perform the appropriate hypothesis test to determine if there’s evidence that there is a difference in average page views
for these two websites. Use a significance level of 0.01.
Ho: Mu1 = Mu2
Ha: Mu1 =/ Mu2
A = .01
T* = (6.2-5.2)/sqrt((2.1^2/32)+(1.2^2/36)) = 2.37
T = 2.75
Fail to reject the null
Comparing 2.37 to the cutoff of 2.75 there is not enough evidence to suggest that there is a difference in average length of a customer service call for the two centers. b.
Check to make sure the assumptions needed to answer the previous questions are reasonable.
Independence: Each call should be independent of one another within their own group and center A and center B should be independent of one another as well. Randomization: States random sample
10% condition: 36 and 32 should be less than 10% of all calls to the center
Nearly Normal: 36>30 and 32>30 so the distribution should be normal
6.
(2 points) State a Type I and Type II error in the context of the previous exercise.
Type II error could occur because we failed to reject the null when it could be false. 7.
(3 points) In order to estimate the true difference in proportion of adults between the ages of 27 and 35 with a college degree across two cities, 400 adults were randomly sampled in Pittsburgh and 350 were sampled in Philadelphia. Of those sampled in Pittsburgh, 30% said they had a college degree, whereas 32% of those sampled in Philadelphia said the same. Calculate and interpret the 95% confidence interval estimate for the true difference in proportion of those with college degrees for these two cities.
Ho: p1 = p2
Ha: p1 =/ p2
N1 = 400
N2 = 350
P(hat)1 = .3
P(hat)2 = .32
(.32-.3) +- (1.96)(sqrt(((.32*.68)/350)+((.3*.7)/400)) = (-.0464, .0864)
We are 95% confident that the true difference in proportion of those with college degrees for these two cities is between -.0464 and .0864.
8.
(4 points) For the previous example perform the appropriate hypothesis test to determine if there’s evidence that there is a difference in proportion of those with college degrees for these two cities. Use a significance level of 0.05. Independence: Those with college degrees should be independent of one another as well as those with degrees from the two cities. Randomization: States random sample
10% condition: 350 and 400 are less than 10% of all people in their respected cities with degrees. Nearly Normal: 350 and 400 are greater than 30 so the distribution is nearly normal. Ho: p1 = p2
Ha: p1 =/ p2
N1 = 400
N2 = 350
P(hat)1 = .3
P(hat)2 = .32
A = .05
Z = (.32-.3)/sqrt(((.32*.68)/350)+((.3*.7)/400)) = .59 = .7224
Fail to Reject the null
Considering a p value of .7224 at a significance level of .05/2 there is not enough evidence to suggest there is a difference in
those with college degrees from these two cities.
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