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 = ?

Step-by-step explanation:

Given that:

The sample mean

The standard deviation = 9

Population mean = 20

Null hypothesis:

Alternative hypothesis:

(a)

When Sample size = 10

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

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

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.

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure

the answer is 13 cm