In: Statistics and Probability
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QUESTION 1
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1 POINT
A fitness center claims that the mean amount of time that a person spends at the gym per visit is 33 minutes. Identify the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter μ.
Select the correct answer below:
H0: μ≠33; Ha: μ=33
H0: μ=33; Ha: μ≠33
H0: μ≥33; Ha: μ<33
H0: μ≤33; Ha: μ>33
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QUESTION 2
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1 POINT
The answer choices below represent different hypothesis tests. Which of the choices are right-tailed tests? Select all correct answers.
Select all that apply:
H0:X≥17.1, Ha:X<17.1
H0:X=14.4, Ha:X≠14.4
H0:X≤3.8, Ha:X>3.8
H0:X≤7.4, Ha:X>7.4
H0:X=3.3, Ha:X≠3.3
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QUESTION 3
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1 POINT
Find the Type II error given that the null hypothesis, H0, is: a building inspector claims that no more than 15% of structures in the county were built without permits.
Select the correct answer below:
The building inspector thinks that no more than 15% of the structures in the county were built without permits when, in fact, no more than 15% of the structures really were built without permits.
The building inspector thinks that more than 15% of the structures in the county were built without permits when, in fact, more than 15% of the structures really were built without permits.
The building inspector thinks that more than 15% of the structures in the county were built without permits when, in fact, at most 15% of the structures were built without permits.
The building inspector thinks that no more than 15% of the structures in the county were built without permits when, in fact, more than 15% of the structures were built without permits.
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QUESTION 4
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1 POINT
Suppose a chef claims that her meatball weight is less than 4 ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a 10% significance level, to persuade them. She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experience that the standard deviation for her meatball weight is 0.5 ounces.
What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?
Provide your answer below:
$$Test statistic =−2.24
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QUESTION 5
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1 POINT
What is the p-value of a right-tailed one-mean hypothesis test, with a test statistic of z0=1.74? (Do not round your answer; compute your answer using a value from the table below.)
z1.51.61.71.81.90.000.9330.9450.9550.9640.9710.010.9340.9460.9560.9650.9720.020.9360.9470.9570.9660.9730.030.9370.9480.9580.9660.9730.040.9380.9490.9590.9670.9740.050.9390.9510.9600.9680.9740.060.9410.9520.9610.9690.9750.070.9420.9530.9620.9690.9760.080.9430.9540.9620.9700.9760.090.9440.9540.9630.9710.977
Provide your answer below:
0.0410
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QUESTION 6
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1 POINT
Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below.
Select the correct answer below:
Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04.
Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04.
Reject the null hypothesis because the value of z is negative.
Reject the null hypothesis because |−1.75|>0.04.
Do not reject the null hypothesis because |−1.75|>0.04.
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QUESTION 7
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1 POINT
A recent study suggested that 81% of senior citizens take at least one prescription medication. Amelia is a nurse at a large hospital who would like to know whether the percentage is the same for senior citizen patients who go to her hospital. She randomly selects 59 senior citizens patients who were treated at the hospital and finds that 49 of them take at least one prescription medication. What are the null and alternative hypotheses for this hypothesis test?
Select the correct answer below:
{H0:p=0.81Ha:p>0.81
{H0:p≠0.81Ha:p=0.81
{H0:p=0.81Ha:p<0.81
{H0:p=0.81Ha:p≠0.81
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QUESTION 8
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1 POINT
A researcher claims that the proportion of cars with manual transmission is less than 10%. To test this claim, a survey checked 1000 randomly selected cars. Of those cars, 95 had a manual transmission.
The following is the setup for the hypothesis test:
{H0:p=0.10Ha:p<0.10
Find the test statistic for this hypothesis test for a proportion. Round your answer to 2 decimal places.
Provide your answer below:
$$Test_Statistic=−0.53
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QUESTION 9
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1 POINT
A medical researcher claims that the proportion of people taking a certain medication that develop serious side effects is 12%. To test this claim, a random sample of 900 people taking the medication is taken and it is determined that 93 people have experienced serious side effects. .
The following is the setup for this hypothesis test:
H0:p = 0.12
Ha:p ≠ 0.12
Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places.
The following table can be utilized which provides areas under the Standard Normal Curve:
In hypothesis, usually the claim to be tested is taken as the alternative hypothesis and the negation of the claim (always with an 'Equal to' sign) is considered as the null hypothesis. And in cases where the claim itself is that a parameter is equal to a hypothesized value, then the claim (with the equal to sign) is considered null and the alternative would be left, right or non-directional as required.
1. Here a fitness center claims that the mean amount of time that a person spends at the gym per visit is 33 minutes and no direction for the claim is specified, hence the correct option would be:
H0: μ=33; Ha: μ≠33
2. We may identify a right-tailed test by looking for our alternative hypothesis statement for a greater than (>) symbol. i.e the inequality would point to the right. Hence, the correct options would be:
H0:X≤3.8, Ha:X>3.8
H0:X≤7.4, Ha:X>7.4
3. Type II error occurs when we accept a false null hypothesis. i.e
H0: A building inspector claims that no more than 15% of structures in the county were built without permits
is false.
i.e In fact, more than 15% of structures in the county were built without permits.
Here, for a type II error to occur, the null H0 must be false, but we would still accept the claim. Hence, the correct option would be:
The building inspector thinks that no more than 15% of the structures in the county were built without permits (Accept H0) when, in fact, more than 15% of the structures were built without permits (when H0 is false).
4.
She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experience that the standard deviation for her meatball weight is 0.5 ounces.
To test: H0: μ≥4; Ha: μ<4 at α=0.1 (significance level)
Given: n = 14,
The appropriate statistical test to test the above hypothesis would be a one-sample t-test with test statistic given by:
5. For Z = 1.74,
Since, the normal table gives only the left tail probabilities,
P-value = P(Z > 1.74) = 1 - :
= 1 - 0.95907
= 0.04093
6. We reject the null hypothesis when:
- The test statistic (Z) lies in the rejection region (Z < Z critical)
- The p-value is less than the fixed level of significance (alpha)
Hence, the correct option would be:
Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04.
7. Here the claim is that 81% of senior citizens take at least one prescription medication and no direction for the claim is specified, hence the correct option would be:
H0:p=0.81Ha:p≠0.81
8. To test:
H0: p = 0.10 Vs H0: p < 0.10
The appropriate statistical test to test the above hypothesis would be a One sample test for Proportion with test statistic given by:
Substituting the given values,
= - 0.527
9. To test:
H0:p = 0.12
Ha:p ≠ 0.12
The appropriate statistical test to test the above hypothesis would be a One sample test for Proportion with test statistic given by:
Substituting the given values,
= - 1.57
For Z = -1.57,
Since, the normal table gives only the left tail probabilities,
P-value = P(|Z| > -1.57) = 1 - :
= 1 - 0.05821
= 0.942