Test the following claim Identify the null hypothesis altern

Test the following claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

A simple random sample of credit rating scores is obtained, and the scores are listed below. The mean credit score was reported to be 680. Assuming the population is normally distributed and the standard deviation of all credit scores is known to be 57.3, use a 0.05 significance level to test the claim these sample credit scores come from a population with a mean equal to 680.

714 750 663 789 818 779 699 836 755 834 691 801

*What are the null and alternative hypothesis?

*What is the value of the test statistic?

z=___

The P-value is ____ (Round to 4 decimla places)

*Fail to reject or reject Ho?

*There (Is / Is not) sufficient evidence to warrant rejection of the claim that these sample credit scores are from a population with a mean credit score equal to 680.

Solution

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Test the following claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.

A simple random sample of credit rating scores is obtained, and the scores are listed below. The mean credit score was reported to be 678. Assuming the population is normally distributed and the standard deviation of all credit scores is known to be 58.1, use a 0.05 significance level to test the claim these sample credit scores come from a population with a mean equal to 678.

713 751 664 789 819 779 699 837 750 834 692 800

what are the null and alternative hypothesis?

What is the value of the test statistic?

The P-value is ?

Fail to reject or reject Ho?

Is there sufficient evidence to warrant rejection of the claim that these sample credit scores are from a population with a mean credit score equal to 678.

Answer

Null Hypothesis: ? = 678

Alternate Hypothesis: ? not = 678

Test Statistic: Z = (x^- - ?)/(?/?n) = 4.92387 ( from TI-83) (you can calculate the z score using this formula and a scientific calculator.)

Read p-value from Z-table : p-value = 0.000000084972 ( from TI-83) ; (from table it is <<<<0.0005)

p-value <<<< ? ==> we reject H_0.

There is not sufficient evidence to warrant the rejection of the claim that this sample is from the same population for which the mean credit score is 678.