Question

                          Hypothesis Test and Confidence Interval for One Mean

1. A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals.                                Data: 4; 2; 1; 3; 7; 2; 9; 4; 6; 6; 7; 0; 5; 6; 4; 2; 1; 3; 4; 2
a. At the α = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year?

α = Ho: Ha:	T – Test inpt: Data Stats µ0: List: Freq:1 µ ≠ µ0< µ0 > µ0 Calculate		T – Test µ0: t = p= x: sx : n:	Decision: Compare the          p-value with alpha and give your decision.
Conclusion:

b. Find a 95 % confidence interval of the true mean years visits of a woman to her doctor.

TInterval Inpt: Data Stats x: sx : n: C-Level: . Calculate	T-Interval (             ,            ) x: sx : n:
Conclusion: We are 95% Confident……

Answer 1

Here Mean

Standard deviation s=

a. Here Alpha 0.05

As it is asked sample mean is higer than 5.8 visits per year   and

Now t-value

=

now p-value

P value = TDIST (t statistics, df, tails)=0.0010

As p-value is smaller than alpha=0.05 we reject the null hypothesis.

b.For 95% confidence interval t-value=

TINV(0.05,19)=

t-value=2.093

Margin or error

Margin of error	1.114
Confidence interval		2.786	5.014

We are 95% confident that population mean will lie between 2.786 and 5.014.