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1. Identify the null hypothesis H0 and the alternative hypothesis

Customer Question

1. Identify the null hypothesis H0 and the alternative hypothesis H1. Use p for a claim about a proportion.
A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO, p, is less than 1 in every one thousand.
A) H0: p = 0.001
H1: p = 0.001
B) H0: p = 0.001
H1: p > 0.001
C) H0: p = 0.001
H1: p < 0.001
D) H0: p = 0.001
H1: p = 0.001
2.Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.
An entomologist writes an article in a scientific journal which claims that fewer than 20 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms.
A) There is not enough information to form a conclusion.
B) There is sufficient evidence to support the claim that the true proportion is less than 20 in ten thousand.
C) A hypothesis test is not the correct test for these conditions.
D) There is not sufficient evidence to support the claim that the true proportion is less than 20 in ten thousand.
3.State the decision rule to test the null hypothesis.
496 people were asked if they were satisfied with their jobs. 39% said they were.
H1: > 0.26. Confidence level: 99%.
A) Reject H0 if the test statistic is > 2.33.
B) Reject H0 if the test statistic is > 2.575.
C) Reject H0 if the test statistic is < 2.085.
D) Reject H0 if the test statistic is > 2.05.

4.Compute the test statistic used to test the null hypothesis.
346 people were asked if they were satisfied with their jobs. 58% said they were.
H1: p > 0.32. Confidence level: 98%.
A) 0.557 B) 10.368 C) 2.612 D) 4.125
5.Use the P-value method to test the given hypothesis and state the P-value.
A nationwide study of American homeowners revealed that 64% have one lawn mower. A lawn equipment manufacturer feels the estimate is wrong for homeowners in Omaha. Test the claim that the number of homeowners in Omaha that have one lawn mower is not 64%. Use a survey of 498 homes (in Omaha) that yields 332 with one lawn mower. Use a level of significance of 0.05.
A) There is not sufficient evidence to support the claim that the proportion of lawn mowers is different than 0.64. The P-value is 0.2150
B) There is not sufficient evidence to support the claim that the proportion of lawn mowers is different than 0.64. The P-value is 0.1075.
C) There is sufficient evidence to support the claim that the proportion of lawn mowers is different than 0.64. The P-value is 0.1075.
D) There is sufficient evidence to support the claim that the proportion of lawn mowers is different than 0.64. The P-value is 0.2150.
6.Find the P-value for the indicated hypothesis test.
A manufacturer claims that fewer than 6% of its fax machines are defective. In a random sample of 97 such fax machines, 5% are defective. Find the P-value for a test of the manufacturer's claim. Assume the level of significance is 0.05.
A) 0.1591 B) 0.1736 C) 0.3264 D) 0.3409

7.Find the P-value for the indicated hypothesis test.
Find the P-value for a test of the claim that more than 50% of the people following a particular diet will experience increased energy. Of 100 randomly selected subjects who followed the diet, 47 noticed an increase in their energy level. Assume the level of significance is 0.05.
A) 0.2257 B) 0.2743 C) 0.5486 D) 0.7257

8.Using the Classical Approach, determine the decision criterion for rejecting the null hypothesis in the given hypothesis test; i.e., describe the values of the test statistic that would result in rejection of the null hypothesis.
H0: s = 37.4
H1: s < 37.4
n = 28
a = 0.05
A) Reject H0 if test statistic > 16.151.
B) Reject H0 if test statistic > 16.928.
C) Reject H0 if test statistic < 16.151.
D) Reject H0 if test statistic < 14.573.
9.Compute the value of an appropriate test statistic for the given hypothesis test and sample data. Assume that the population is normally distributed and that the sample has been randomly selected.
H0: s = 3.3
H1: s > 3.3
n = 13
s = 7.2
a = 0.05
A) X2 = 57.124
B) X2 = 804.496
C) X2 = 26.182
D) X2 = 2.521
10.Use the traditional method to test the given hypothesis and state the critical value.

In one town, monthly incomes for men with college degrees are found to have a standard deviation of $650. Use a 0.01 significance level to test the claim that for men without college degrees in that town, incomes have a higher standard deviation. A random sample of 22 men without college degrees resulted in incomes with a standard deviation of $922.
A) There is sufficient evidence to support the claim that incomes of men without college degrees have a standard deviation greater than $650 and the critical value is 38.932
B) There is not sufficient evidence to support the claim that incomes of men without college degrees have a standard deviation greater than $650 and the critical value is 38.932
C) There is sufficient evidence to support the claim that incomes of men without college degrees have a standard deviation greater than $650 and the critical value is 41.404
D) There is not sufficient evidence to support the claim that incomes of men without college degrees have a standard deviation greater than $650 and the critical value is 41.404
Submitted: 6 years ago.
Category: Math Homework
Expert:  rajeevanpillai replied 6 years ago.
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Customer: replied 6 years ago.
how to solve question 3?
Expert:  rajeevanpillai replied 6 years ago.
H0: p = 0.26
H1: p > 0.26


If Z is the standard normal variable, we can find from tables that
P( Z > 2.33) = 0.01 and P(Z < 2.33) = 0.99
The critical (Rejection) region for the test is Z > 2.33
The decision rule is to reject the null hypothesis if Z > 2.33

The value of Z can be computed as
Z = (0.39-0.26)/sqrt(0.26*(1-0.26)/496) =6.6005

This value of Z satisfies Z > 2.33

So the null hypothesis is rejected.

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