Question

In a sample of credit card holders the mean monthly value of credit card purchases was $ 400 and the sample variance was 67 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate.

(a) Suppose the sample results were obtained from a random sample of 12 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.

(  ,  )

(b) Suppose the sample results were obtained from a random sample of 21 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.

(  ,  )
Before you answer (b) consider whether the confidence interval will be wider than or narrower than the confidence interval found for (a). Then check that your answer verifies this.

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A random sample of 100100 observations from a population with standard deviation 6.366.36 yielded a sample mean of 91.191.1.

1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the following:
(a) Test statistic ==
(b) P - value:
(c) The conclusion for this test is:

A. There is insufficient evidence to reject the null hypothesis
B. Reject the null hypothesis
C. None of the above

2. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ≠90μ≠90 using α=.05α=.05, find the following:
(a) Test statistic ==
(b) P - value:
(c) The conclusion for this test is:

A. There is insufficient evidence to reject the null hypothesis
B. Reject the null hypothesis
C. None of the above

Answer 1

a) DF = 12 - 1 = 11

At 95% confidence level, the critical value is t^* = 2.201

The 95% confidence interval is

It *

29/ * [0%Z+ 00Ꭽ =

= 400 +5.20

= 394.80. 405.20

b) DF = 21 - 1 = 20

At 95% confidence level, the critical value is t^* = 2.086

The 95% confidence interval is

It *

= 400 + 2.086 + V

= 400 +3.73

= 396.27, 403.73

The confidence interval in part(b) will be narrower than the confidence interval in part(a) .

1) a) The test statistic is

urlo

  

91.1 - 90 6.36/1100

= 1.73

b) P-value = P(Z > 1.73)

               = 1 - P(Z < 1.73)

               = 1 - 0.9582

               = 0.0418

c) Since the P-value is less than the significance level(0.0418 < 0.05), so we should reject the null hypothesis.

Option - B) Reject the null hypothesis.

2) The test statistic is

urlo

  

91.1 - 90 6.36/1100

= 1.73

b) P-value = 2 * P(Z > 1.73)

               = 2 * (1 - P(Z < 1.73))

               = 2 * (1 - 0.9582)

               = 0.0836

c) Since the P-value is greater than the significance level(0.0836 > 0.05), so we should not reject the null hypothesis.

Option - A) There is insufficient evidence to reject the null hypothesis.