Question

Consider the following hypotheses:

H₀: μ = 36
H_A: μ ≠ 36

The population is normally distributed. A sample produces the following observations: (You may find it useful to reference the appropriate table: z table or t table)

38	36	38	41	38	38	39

Click here for the Excel Data File
  

a. Find the mean and the standard deviation. (Round your answers to 2 decimal places.)

b. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

c. Find the p-value.

• p-value < 0.01

• p-value 0.10
• 0.05 p-value < 0.10
• 0.02 p-value < 0.05
• 0.01 p-value < 0.02

d. At the 5% significance level, what is the conclusion?

• Do not reject H₀ since the p-value is smaller than α.

• Do not reject H₀ since the p-value is greater than α.

• Reject H₀ since the p-value is smaller than α.

• Reject H₀ since the p-value is greater than α.

e.  Interpret the results αα = 0.05.

• We cannot conclude that the sample mean differs from 36.

• We conclude that the population mean differs from 36.

• We cannot conclude that the population mean differs from 36.

• We conclude that the sample mean differs from 36.

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

H0: µ = 36 versus Ha: µ ≠ 36

This is a two tailed test.

Part a

From given data, we have

µ = 36

Xbar = 38.29

S = 1.50

n = 7

df = n – 1 = 6

α = 0.05

Critical value = - 2.4469 and 2.4469

(by using t-table or excel)

Part b

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

t = (38.28571429 – 36)/[ 1.496026483/sqrt(7)]

t = 4.0423

Part c

P-value = 0.0068

(by using t-table)

p-value < 0.01

Part d

P-value < α = 0.05

So, we reject the null hypothesis

Reject H0 since the p-value is smaller than α.

Part e

We conclude that the population mean differs from 36.