Question

In: Statistics and Probability

Use the given information to find the p-value. also use a0.05 significance level and state the...

Use the given information to find the p-value. also use a0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail tor rejected the null hypothesis)

With H1 : p < 3/5; the test statistic is z = -1.68

A: 0.093 ; fail to reject the null hypothesis
B: 0.0465 ; fail to reject the null hypothesis
C: 0.0465 ; reject the null hypothesis
D: 0.09535 ; fail to reject the null hypothesis

With H1 : p > 0.383, the test statistic is z = 0.41

A: 0.6591 ; fail to reject the null hypothesis
B: 0.3409 ; fail to reject the null hypothesis
C: 0.3490 ; reject the null hypothesis
D: 0.6818 ; reject the null hypothesis

Solutions

Expert Solution

Solution:

Given:

Significance level  = 0.05

Part a)

H1 : p < 3/5; the test statistic is z = -1.68

Since H1 is < type, this is left tailed test.

Thus for left tailed test , p-value is:

p-value = P(Z < z test statistic)

p-value = P(Z < -1.68)

Look in z table for z = -1.6 and 0.08 and find corresponding area.

P( Z<-1.68) = 0.0465

p-value = P(Z < -1.68)

p-value = 0.0465

Decision Rule:
Reject null hypothesis H0, if p-value < 0.05 level of significance, otherwise we fail to reject H0

Since p-value = 0.0465 < 0.05 level of significance, we reject null hypothesis H0.

Thus correct answer is:

C: 0.0465 ; reject the null hypothesis

Part b)

H1 : p > 0.383, the test statistic is z = 0.41

Since H1 is > type , this is right tailed test.

For right tailed test , p-value is:

p-value = P(Z > z test statistic)

p-value = P(Z > 0.41)

p-value = 1 - P(Z < 0.41)

Look in z table for z = 0.4 and 0.01 and find corresponding area.

P( Z < 0.41) = 0.6591

thus

p-value = 1 - P(Z < 0.41)

p-value = 1 -0.6591

p-value = 0.3409

Since p-value = 0.3409 >  0.05 level of significance, we fail to reject H0.

Thus correct answer is:

B: 0.3409 ; fail to reject the null hypothesis


Related Solutions

Use the given information to find the ​p-value. ​Also, use a 0.05 significance level and state...
Use the given information to find the ​p-value. ​Also, use a 0.05 significance level and state the conclusion about the null hypothesis​ (reject the null hypothesis or fail to reject the null​ hypothesis). With H1​: p> ​0.554, the test statistic is z=1.34.
A) Assume that the significance level is α=0.05. Use the given information to find the​ P-value...
A) Assume that the significance level is α=0.05. Use the given information to find the​ P-value and the critical​ value(s). The test statistic of z = −1.09 is obtained when testing the claim that p<0.2. B) Identify the type I error and the type II error that corresponds to the given hypothesis. The proportion of settled medical malpractice suits is 0.29. C) A survey of 1,562 randomly selected adults showed that 579 of them have heard of a new electronic...
Use .05 level of significance Find the P-Value- Find the test statistic Zo- three sixty two...
Use .05 level of significance Find the P-Value- Find the test statistic Zo- three sixty two / (eleven hundred) successful three ninety six / (eleven hundred) successful Did proportion change? It just says one was 50 years ago the 362 and the next one was is today. thats all it says, theres no sd
From the given information in each case below, use technology to find the P-value for a...
From the given information in each case below, use technology to find the P-value for a chi-square test and give the conclusion for a significance level of α = 0.01. (Use technology to calculate the P-value. Round your answers to three decimal places.) (a) χ2 = 7.4, df = 2 P-value = State the conclusion. Reject H0.Fail to reject H0.    (b) χ2 = 13.1, df = 6 P-value = State the conclusion. Reject H0.Fail to reject H0. = (c)...
Use the information below to determine whether.  State the statistical hypotheses and report the P-value.  Find the 95%...
Use the information below to determine whether.  State the statistical hypotheses and report the P-value.  Find the 95% confidence intervals. Sample 1: N1=48, X1= 4.99, ss1= 13.2 Sample 2: N2= 32, X2= 5.83, ss2= 29.2
using 0.01 level of significance use the information given to test whether a person's ability in...
using 0.01 level of significance use the information given to test whether a person's ability in math is independent of his or her interest in calculus. Ability in math Low Average High   Low 63 42 15 Average 58 61 31 High 14 47 29 Interest in calculus - left
Question #3 Use P-Value method and 3% significance level, test the claim that the true mean...
Question #3 Use P-Value method and 3% significance level, test the claim that the true mean weight loss produced by the three exercise programs have the same mean. Assume that the populations are normally distributed with the same variance. Exercise A Exercise B Exercise C 5.6 6.8 9.3 8.1 4.9 6.2 4.3 3.1 5.8 9.1 7.8 7.1 7.1 1.2 7.9
A null hypothesis is not rejected at a given level of significance. As the assumed value...
A null hypothesis is not rejected at a given level of significance. As the assumed value of the mean gets further away from the true population mean, the Type II error will
Write a paragraph or more to explain what p-value and significance level are, and what they...
Write a paragraph or more to explain what p-value and significance level are, and what they are for?
At a .01 significance level with a sample size of 18, find the critical value for...
At a .01 significance level with a sample size of 18, find the critical value for the correlation coefficient of 1.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT