Answer to Question #100083 in Statistics and Probability for Juliet Beglaryan

Question

The average amount of time a woman spends on house work is claimed to be normally distributed at 7 hr/week and populaton standard deviation is 1.5. If a researcher measures the average for 45 sampled women to be 8.5 hours/week. 1) what is the hypothesis (H0 and H1). 2) Verify necessary data condtiions for a z statistics and calculate the z statisitcs. 3) Are the results satistically significant can we reject the null or not? 4) For what values of x bar do we reject the null?

Answer 1

We have, $\mu=7,\sigma=1.5$

Sample size, $n=45$ and sample mean, $\bar{X}=8.5$

1)

we can assume that

Null hypothesis, $H_0:\mu=7$

Alternate hypothesis, $H_1: \mu>7$

2)

Since the sample is randomly selected from a normal distribution population whose standard deviation is known, and also the size of the sample is greater than 30, we can apply the Z-test.

Z-statistics $=\frac{(\bar{X}-\mu_0)}{\sigma/\sqrt{n}}$

Putting up all the values, we get:

$z=\frac{8.5-7}{1.5/\sqrt{45}}$

$=\frac{1.5}{1.5/6.70}=6.70$

3)

Since it is one tail test, the P-value for this test will be $P(Z>6.70)$ , which from the normal distribution table is approximately equal to zero.

So, we can reject the null hypothesis at any value of alpha which is slightly greater than Zero.

4)

For the given value of $\bar{X}$ , we can reject the null hypothesis for any value of alpha.

Or suppose $\alpha=0.01$ , then to reject the null hypotheses, p-value should be less than alpha, that means $P(Z>z)<0.01$ or $P(Z<z)>0.99$

Then from the normal distribution table, value of z should be greater than 2.326.

So,

$\frac{(\bar{X}-\mu_0)}{\sigma/\sqrt{n}}>2.326$

$=\frac{(\bar{X}-7)}{1.5/\sqrt{45}}>2.326$

$=\frac{(\bar{X}-7)}{0.224}>2.326$

$=(\bar{X}-7)>0.520$

$=\bar{X}>0.520+7$

$or,\bar{X}>7.520$