**Consider the following hypotheses:**

**
**

a.

Critical value(s)
±

b-1.

Test statistic

b-2.
What is the conclusion at α = 0.05?
Reject H0 since the value of the test statistic is greater than the critical value.
Reject H0 since the value of the test statistic is smaller than the critical value.
Do not reject H0 since the value of the test statistic is greater than the critical value.
Do not reject H0 since the value of the test statistic is smaller than the critical value.

c.

Critical value(s)
±

d-1.

Test statistic

d-2. What is the conclusion at α = 0.10? Reject H0 since the value of the test statistic is not less than the negative critical value. Reject H0 since the value of the test statistic is less than the negative critical value. Do not reject H0 since the value of the test statistic is not less than the negative critical value. Do not reject H0 since the value of the test statistic is less than the negative critical value. 6.

Test statistic

Test statistic

Confidence Level
Confidence Interval
90%
%
to
%
99%
%
to
%
8.

a.

Confidence interval
to

b.

a.

Confidence interval
to

b.

H0: μ = 120

HA: μ ≠ 120
The population is normally distributed with a population standard deviation of 46. Use Table 1.

Use a 5% level of significance to determine the critical value(s) of the test. (Round your answer to 2 decimal places.)

Calculate the
value of the test statistic with = 132 and n = 50. (Round intermediate
calculations to 4 decimal places. Round your answer to 2 decimal
places.)

Use a 10% level of significance to determine the critical value(s) of the test. (Round your answer to 2 decimal places.)

Calculate the
value of the test statistic with = 108 and n = 50. (Negative value
should be indicated by a minus sign. Round intermediate calculations to 4
decimal places. Round your answer to 2 decimal places.)

d-2. What is the conclusion at α = 0.10? Reject H0 since the value of the test statistic is not less than the negative critical value. Reject H0 since the value of the test statistic is less than the negative critical value. Do not reject H0 since the value of the test statistic is not less than the negative critical value. Do not reject H0 since the value of the test statistic is less than the negative critical value. 6.

In order to
conduct a hypothesis test of the population mean, a random sample of 24
observations is drawn from a normally distributed population. The
resulting mean and the standard deviation are calculated as 4.8 and 0.8,
respectively. Use Table 2.

Use the p-value approach to conduct the following tests at α = 0.05.

H0: μ ≤ 4.5 against HA: μ > 4.5

a-1.

Calculate the
value of the test statistic. (Round intermediate calculations to 4
decimal places and your answer to 2 decimal places.)

a-2.

Approximate the p-value.

0.05 < p-value < 0.10
0.005 < p-value < 0.01
0.025 < p-value < 0.05
a-3.

What is the conclusion?

Reject H0 since the p-value is less than α.
Reject H0 since the p-value is greater than α.
Do not reject H0 since the p-value is less than α.
Do not reject H0 since the p-value is greater than α.
H0: μ = 4.5 against HA: μ ≠ 4.5

b-1.

Calculate the
value of the test statistic. (Round intermediate calculations to 4
decimal places and your answer to 2 decimal places.)

b-2.

Approximate the p-value.

0.05 < p-value < 0.10
0.025 < p-value < 0.05
0.005 < p-value < 0.01
b-3.

What is the conclusion?

Reject H0 since the p-value is less than α.
Reject H0 since the p-value is greater than α.
Do not reject H0 since the p-value is less than α.
Do not reject H0 since the p-value is greater than α.
In order to
estimate the mean 30-year fixed mortgage rate for a home loan in the
United States, a random sample of 28 recent loans is taken. The average
calculated from this sample is 5.25%. It can be assumed that 30-year
fixed mortgage rates are normally distributed with a standard deviation
of 0.50%. Compute 90% and 99% confidence intervals for the population
mean 30-year fixed mortgage rate. Use Table 1.(Round
intermediate calculations to 4 decimal places, "z" value and final
answers to 2 decimal places. Enter your answers as percentages, not
decimals.)

An article in
the National Geographic News (“U.S. Racking Up Huge Sleep Debt,”
February 24, 2005) argues that Americans are increasingly skimping on
their sleep. A researcher in a small Midwestern town wants to estimate
the mean weekday sleep time of its adult residents. He takes a random
sample of 80 adult residents and records their weekday mean sleep time
as 6.4 hours. Assume that the population standard deviation is fairly
stable at 1.8 hours. Use Table 1.

Calculate a
95% confidence interval for the population mean weekday sleep time of
all adult residents of this Midwestern town. (Round intermediate
calculations to 4 decimal places, "z" value and final answers to 2
decimal places.)

Can we conclude with 95% confidence that the mean sleep time of all adult residents in this Midwestern town is not 7 hours?

Yes, since the confidence interval contains the value 7.
Yes, since the confidence interval does not contain the value 7.
No, since the confidence interval contains the value 7.
No, since the confidence interval does not contain the value 7.
9.
A random sample of
24 observations is used to estimate the population mean. The sample
mean and the sample standard deviation are calculated as 104.6 and 28.8,
respectively. Assume that the population is normally distributed. Use Table 2.

Construct a 90%
confidence interval for the population mean. (Round intermediate
calculations to 4 decimal places, "t" value to 3 decimal places, and
final answers to 2 decimal places.)

Construct a 99%
confidence interval for the population mean. (Round intermediate
calculations to 4 decimal places, "t" value to 3 d

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