Question

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 13 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 12.5.

(a) Is it appropriate to use a Student's *t*
distribution? Explain.

Yes, because the *x* distribution is mound-shaped and
symmetric and *σ* is unknown.No, the *x* distribution
is skewed left. No, the *x*
distribution is skewed right.No, the *x* distribution is not
symmetric.No, *σ* is known.

How many degrees of freedom do we use?

(b) What are the hypotheses?

*H*_{0}: *μ* = 12.5;
*H*_{1}: *μ* > 12.5*H*_{0}:
*μ* < 12.5; *H*_{1}: *μ* =
12.5 *H*_{0}: *μ* =
12.5; *H*_{1}: *μ* <
12.5*H*_{0}: *μ* > 12.5;
*H*_{1}: *μ* = 12.5*H*_{0}:
*μ* = 12.5; *H*_{1}: *μ* ≠ 12.5

(c) Compute the *t* value of the sample test statistic.
(Round your answer to three decimal places.)

* t* =

(d) Estimate the *P*-value for the test.

*P*-value > 0.2500.100 < *P*-value <
0.250 0.050 < *P*-value <
0.1000.010 < *P*-value < 0.050*P*-value <
0.010

(e) Do we reject or fail to reject *H*_{0}?

At the *α* = 0.05 level, we reject the null hypothesis
and conclude the data are statistically significant.At the
*α* = 0.05 level, we reject the null hypothesis and conclude
the data are not statistically
significant. At the *α* = 0.05 level,
we fail to reject the null hypothesis and conclude the data are
statistically significant.At the *α* = 0.05 level, we fail
to reject the null hypothesis and conclude the data are not
statistically significant.

(f) Interpret the results.

There is sufficient evidence at the 0.05 level to reject the null hypothesis.There is insufficient evidence at the 0.05 level to reject the null hypothesis.

Answer #1

**(a) Yes,
because the x distribution is mound-shaped and symmetric
and σ is unknown**

Degree of freedom = n-1 = 16-1 = 15

**(b)
H_{0}: μ = 12.5; H_{1}:
μ ≠ 12.5**

Since it is a two tailed test, we are saying that mean is not equal; it can be less than or greater than 12.5

**(c) t
value**

**(d) P
value**

P value = TDIST (t statistics, df, 2) = TDIST (1, 15, 2) = 0.333

*P*-value >
0.250

**(e)** level of
significance = 0.05 and P value os 0.333

Since P value is greater than level of significance, we fail to reject the Null hypothesis.

**At the
α = 0.05 level, we fail to reject the null hypothesis and
conclude the data are not statistically
significant.**

(f) **There is
insufficient evidence at the 0.05 level to reject the null
hypothesis.**

A random sample of 16 values is drawn from a mound-shaped and
symmetric distribution. The sample mean is 11 and the sample
standard deviation is 2. Use a level of significance of 0.05 to
conduct a two-tailed test of the claim that the population mean is
10.5.
(a) Is it appropriate to use a Student's t
distribution? Explain.
Yes, because the x distribution is mound-shaped and
symmetric and σ is unknown.No, the x distribution
is skewed left. No, the x
distribution...

A random sample of 36 values is drawn from a mound-shaped and
symmetric distribution. The sample mean is 14 and the sample
standard deviation is 2. Use a level of significance of 0.05 to
conduct a two-tailed test of the claim that the population mean is
13.5.
(a) Is it appropriate to use a Student's t
distribution? Explain.
Yes, because the x distribution is mound-shaped and
symmetric and σ is unknown.No, the x distribution
is skewed left. No, the x
distribution...

A random sample of 36 values is drawn from a mound-shaped and
symmetric distribution. The sample mean is 9 and the sample
standard deviation is 2. Use a level of significance of 0.05 to
conduct a two-tailed test of the claim that the population mean is
8.5.
(a) Is it appropriate to use a Student's t
distribution? Explain.
Yes, because the x distribution is mound-shaped and
symmetric and σ is unknown.No, the x distribution
is skewed left. No, the x
distribution...

A random sample of 25 values is drawn from a mound-shaped and
symmetric distribution. The sample mean is 11 and the sample
standard deviation is 2. Use a level of significance of 0.05 to
conduct a two-tailed test of the claim that the population mean is
10.5.
(a) Is it appropriate to use a Student's t distribution?
Explain.
Yes, because the x distribution is mound-shaped and symmetric
and σ is unknown.
No, the x distribution is skewed left.
No, the...

A random sample of 36 values is drawn from a mound-shaped and
symmetric distribution. The sample mean is 8 and the sample
standard deviation is 2. Use a level of significance of 0.05 to
conduct a two-tailed test of the claim that the population mean is
7.5.
(a) Is it appropriate to use a Student's t
distribution? Explain.
Yes, because the x distribution is mound-shaped and
symmetric and σ is unknown.
No, the x distribution is skewed
left.
No, the...

A random sample of 36 values is drawn from a mound-shaped and
symmetric distribution. The sample mean is 11 and the sample
standard deviation is 2. Use a level of significance of 0.05 to
conduct a two-tailed test of the claim that the population mean is
10.5.
(a) Is it appropriate to use a Student's t
distribution? Explain.
Yes, because the x distribution is mound-shaped and
symmetric and σ is unknown.No, the x distribution is
skewed left. No, the x
distribution...

A random sample of 40 binomial trials resulted in 16 successes.
Test the claim that the population proportion of successes does not
equal 0.50. Use a level of significance of 0.05.
(a) Can a normal distribution be used for the p̂
distribution? Explain.
No, nq is greater than 5, but np is less than
5.Yes, np and nq are both greater than
5. No, np is greater than 5, but
nq is less than 5.No, np and nq are both
less...

Let x be a random variable that represents the pH of
arterial plasma (i.e., acidity of the blood). For healthy adults,
the mean of the x distribution is μ = 7.4.† A new
drug for arthritis has been developed. However, it is thought that
this drug may change blood pH. A random sample of 36 patients with
arthritis took the drug for 3 months. Blood tests showed that
x = 8.5 with sample standard deviation s = 2.8.
Use a...

Let x be a random variable that represents the
pH of arterial plasma (i.e., acidity of the blood). For healthy
adults, the mean of the x distribution is μ = 7.4.† A new
drug for arthritis has been developed. However, it is thought that
this drug may change blood pH. A random sample of 31 patients with
arthritis took the drug for 3 months. Blood tests showed that x =
8.4 with sample standard deviation s = 2.6. Use a...

A random sample of 50 binomial trials resulted in 20 successes.
Test the claim that the population proportion of successes does not
equal 0.50. Use a level of significance of 0.05.
(e)
Do you reject or fail to reject H0?
Explain.
At the α = 0.05 level, we reject the null hypothesis and
conclude the data are statistically significant.At the α = 0.05
level, we reject the null hypothesis and conclude the data are not
statistically significant. At the α =...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 3 minutes ago

asked 11 minutes ago

asked 11 minutes ago

asked 14 minutes ago

asked 20 minutes ago

asked 21 minutes ago

asked 37 minutes ago

asked 38 minutes ago

asked 39 minutes ago

asked 42 minutes ago

asked 48 minutes ago