Question

1.
using a significance level of 0.05, find P value

identify Ho & Ha.

show whether left tailed, right or two tailed.

p> .32 , z = -2.41

same for : t= -0.29 , n = 16

2. There is about 200 single people in the united states in
the 1040s, they are randomly selected. the mesn income of the group
of 200 is about 28,000 and has a standard deviaton of
$40,000.

cl : 95%

Answer #1

Question 1

From the given information, we have

H0: p = 0.32 versus Ha: p > 0.32

This is an upper or right tailed test.

Test statistic = z = -2.41

P-value = 0.992024

(by using z-table)

Now for test statistic = t = -0.29

n = 16, df = n - 1 = 15

P-value = 0.612107

(by using t-table)

Question 2

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 28000

S = 40000

n = 200

df = n – 1 = 199

Confidence level = 95%

Critical t value = 1.9720

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 28000 ± 1.9720*40000/sqrt(200)

Confidence interval = 28000 ± 5577.5352

Lower limit = 28000 - 5577.5352 = 22422.46

Upper limit = 28000 + 5577.5352 = 33577.54

Confidence interval = (22422.46, 33577.54)

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).
(a) The test statistic in a right-tailed test is z = 2.00
(b) The test statistic of z = -1.75 is obtained when testing the
claim that p = 1/3

*Please show your work
*Be sure to round z-values to 2 decimal places and round
any probabilities to 4 decimal places
*If doing a hypothesis test, be sure to sure state the
Ho and Ha, you may choose the p-value or rejection region approach
to hypothesis testing, unless otherwise noted in the
question.
*If you need to use the normal approximation, be sure to
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