Question

Each person in a random sample of 228 male teenagers and a random sample of 305 female teenagers was asked how many hours he or she spent online in a typical week. The sample mean and standard deviation were 15.2 hours and 11.3 hours for males and 14.2 and 11.7 for females. (Use a statistical computer package to calculate the P-value. Use μmales − μfemales. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

t = df = P =

Answer 1

Let's use TI-84 Plus calculator.

Click on STAT >>>TESTS>>> 4: 2-SampTTest...>>>Stats

: 15.2

Sx : 11.3

n1 : 228

: 14.2

Sy : 11.7

n2 : 305

( Highlight ) and ENTER

pooled: Yes (Highlight Yes and ENTER )

Then highlight Calculate and ENTER

Look the following image.

So we get the following output.

From the above statistic, we get

t = 0.99

df = 531

P = 0.322

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.322 > 0.05 so we used second rule.

That is we fail to reject null hypothesis ( H₀ ).