Question

A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 100 greeting cards indicates a mean value of ​$2.53 and a standard deviation of ​$0.41 . Complete parts​ (a) and​ (b).

A. Is there evidence that the population mean retail value of the greeting cards is different from $2.50 ​? ​(Use a 0.10 level of​ significance.)

State the null and alternative hypothesis. Identify the critical value(s). Determine the test statistic. What is the conclusion?

B. Determine the p-value and interpret its meaning

Answer 1

Solution:

Part a

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

The null and alternative hypotheses are given as below:

Null hypothesis: H₀: the population mean retail value of the greeting cards is $2.50.

Alternative hypothesis: H_a: the population mean retail value of the greeting cards is different from $2.50.

H₀: µ = 2.50 versus H_a: µ ≠ 2.50

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 2.50

Xbar = 2.53

S = 0.41

n = 100

df = n – 1 = 99

α = 0.10

Critical value = -1.6604 and 1.6604

(by using t-table or excel)

t = (2.53 – 2.50)/[0.41/sqrt(100)]

t = 0.7317

Test statistic is lies within critical values, so we do not reject the null hypothesis.

There is not sufficient evidence to conclude that the population mean retail value of the greeting cards is different from $2.50.

Part b

P-value = 0.4661

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis