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Question 1
1/1 points
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, H 0 , and the alternative hypothesis, Ha , in terms of the parameter μ. That is correct! H 0 : μ ≠ 33 ; Ha : μ =3 3 H 0 : μ =3 3 ; Ha : μ ≠ 33 H 0 : μ ≥ 33 ; Ha : μ <3 3 H 0 : μ ≤ 33 ; Ha : μ >3 3 Answer Explanation Correct answer: H 0 : μ =3 3 ; Ha : μ ≠ 33 Let the parameter μ be used to represent the mean. The null hypothesis is always stated with some form of equality: equal (=), greater than or equal to (≥), or less than or equal to (≤).
Therefore, in this case, the null hypothesis H 0 is μ =3 3. The alternative hypothesis is contradictory to the null hypothesis, so Ha is μ ≠ 33.
H 0 : X ≤ 3. 8 , Ha : X >3. 8
H 0 : X ≤ 7. 4 , Ha : X >7. 4
H 0 : X =3. 3 , Ha : X ≠ 3. 3
Answer Explanation Correct answer: H 0 : X ≤ 3. 8 , Ha : X >3. 8 H 0 : X ≤ 7. 4 , Ha : X >7. 4 Remember the forms of the hypothesis tests.
- Right-tailed: H 0 : X ≤ X 0 , Ha : X > X 0.
- Left-tailed: H 0 : X ≥ X 0 , Ha : X < X 0.
- Two-tailed: H 0 : X = X 0 , Ha : X ≠ X 0. So in this case, the right-tailed tests are:
- H 0 : X ≤ 7. 4 , Ha : X >7. 4
- H 0 : X ≤ 3. 8 , Ha : X >3. 8
1/1 points
Find the Type II error given that the null hypothesis, H 0 , is: a building inspector claims that no more than 15 %of structures in the county were built without permits. That is correct! 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. Answer Explanation Correct answer: 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. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this
case, the Type II error is when the building inspector thinks that no more than 15 %of the
Answer Explanation Correct answers:
- Test statistic = minus 2 point 2 4 $\text{Test statistic = }-2.24$
The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean weight, x ¯=3. 7. The sample the chef uses is 14 meatballs, so n =1 4. She knows the standard deviation of the meatballs, σ =0. 5. Lastly, the chef is comparing the population mean weight to 4 ounces. So, this value (found in the null and alternative hypotheses) is μ 0. Now we will substitute the values into the formula to compute the test statistic: z 0 = x ¯− μ 0 σn √=3. 7 − 40. 514 √≈− 0. 30. 134 ≈− 2. 24 So, the test statistic for this hypothesis test is z 0 =− 2. 24.
Question 5
1/1 points
What is the p -value of a right-tailed one-mean hypothesis test, with a test statistic of z 0 =1. 74? (Do not round your answer; compute your answer using a value from the table below.) z 1. 51. 61. 71. 81. 90. 00 0. 9330. 9450. 9550. 9640. 971 0. 01 0. 9340. 9460. 9560. 9650. 972 0. 02 0. 9360. 9470. 9570. 9660. 973 0. 03 0. 9370. 9480. 9580. 9660. 973 0. 04 0. 93
- 974 0. 05 0. 9390. 9510. 9600. 9680. 974 0. 06 0. 9410. 9520. 961
- 975 0. 07 0. 9420. 9530. 9620. 9690. 976 0. 08 0. 9430. 9540. 9620. 9700. 976 0
0 point 0 4 1$$ 0 point 0 4 1 - correct Answer Explanation Correct answers:
The p -value is the probability of an observed value of z= 1. 74 or greater if the null hypothesis is true, because this hypothesis test is right-tailed. This probability is equal to the area under the Standard Normal curve to the right of z =1. 74. A standard normal curve with two points labeled on the horizontal axis. The mean is labeled at 0.00 and an observed value of 1.74 is labeled. The area under the curve and to the right of the observed value is shaded. Using the Standard Normal Table, we can see that the p -value is equal to 0. 959 , which is the area to the left of z =1. 74. (Standard Normal Tables give areas to the left.) So, the p -value we're looking for is p =1− 0. 959 =0. 041.
Reject the null hypothesis because the value of z is negative. Reject the null hypothesis because |− 1. 75 |>0. 04.
Do not reject the null hypothesis because |− 1. 75 |>0. 04.
Answer Explanation
Correct answer: Do not reject the null hypothesis because the p -value 0. 0401 is greater than the significance level α =0. 04. In making the decision to reject or not reject H 0 , if α > p -value, reject H 0 because the results of the sample data are significant. There is sufficient evidence to conclude that H 0 is an incorrect belief and that the alternative hypothesis, Ha , may be correct. If α ≤ p -value, do not reject H 0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha , may be correct. In this case, α =0. 04 is less than or equal to p =0. 0401 , so the decision is to not reject the null hypothesis.
Question 45
1/1 points
Fill in the following contingency table and find the number of students who both do not play sports AND do not play an instrument. StudentsplaysportsdonotplaysportsTotalplayaninstrument 33 donotplayaninstru ment 69 Total 6267 That is correct!
34$$ 34 - correct
Answer Explanation
Correct answers:
By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 33 added to the unknown number in the middle is 67 , so that unknown number is 34. Continuing in this way, we can fill in the entire table: StudentsplaysportsdonotplaysportsTotalplayaninstrument 273360 donotplayani nstrument 353469 Total 6267129 From this, we can see that the number of students who both do not play sports and do not play an instrument is 34.
Question 46
1/1 points
The answer choices below represent different hypothesis tests. Which of the choices are left- tailed tests? Select all correct answers.
That is correct!
