Question

The sample mean and standard deviation from a random sample of 10 observations from a normal population were computed as ?x=23????? and ?????s=9
Calculate the value of the test statistic (to 3 decimals) and the p-value (to 4 decimals) of the test required to determine whether there is enough evidence to infer at the 5% significance level that the population mean is greater than 20
t = ? p-value = ?
1b) Repeat Part (a) with ????n = 30
t = ? p-value = ?
1c) Repeat Part (a) with n = 50
t = ? p-value = ?

Answer 1

Answer:

Step-by-step explanation:

Given that:

The sample mean $\overline x = 23$

The standard deviation $\sigma$ = 9

Population mean = 20

Null hypothesis:

$H_o: \mu = 20$

Alternative hypothesis:

$H_1 : \mu> 30$

(a)

When Sample size = 10

$Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }$

$=\dfrac{23 -20}{\dfrac{9}{\sqrt{10}} }$

$=\dfrac{3 \times \sqrt{10}}{9 }$

t = 1.0541

Degree of freedom df:

df = n -1

df = 10 - 1

df = 9

P(value) for t = 1.0541 at df = 9:

P(value) = P(Z > 1.0541)

P(value) = 1 - P(< 1.0541)

P(value) = 1 - 0.8403

P(value) = 0.1597

There is no enough evidence to infer at the 5% significance since p-value is greater than the level of significance.

(b) When sample size = 30

$Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }$

$=\dfrac{23 -20}{\dfrac{9}{\sqrt{30}} }$

$=\dfrac{3 \times \sqrt{30}}{9 }$

t = 1.8257

Degree of freedom df:

df = n -1

df = 30 - 1

df = 29

P(value) for t =  1.8257 at df = 29:

P(value) = P(Z > 0.9609)

P(value) = 1 - P(< 0.9609)

P(value) = 1 - 0.9609

P(value) = 0.0391

There is enough evidence to infer that the mean is greater than 20 at the 5%  significance level as the p-value is less than the significance level.

(c) When sample size = 50

$Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }$

$=\dfrac{23 -20}{\dfrac{9}{\sqrt{50}} }$

$=\dfrac{3 \times \sqrt{50}}{9 }$

t = 2.3570

Degree of freedom df:

df = n -1

df = 50 - 1

df = 49

P(value) for t =  2.3570 at df = 49:

P(value) = P(Z > 0.9888)

P(value) = 1 - P(< 0.9888)

P(value) = 1 - 0.9888

P(value) = 0.0112

There is enough evidence to infer that the mean is greater than 20 at the 5%  significance level as the p-value is less than the significance level.