Question

Could someone please help me solve this. The correct answers are shown. I keep getting this question wrong and cannot wrap my head around how to do it. Thank you!

For a? two-tailed hypothesis test at the significance level ??, the null hypothesis H0?: ?=?0 will be rejected in favor of the alternative hypothesis Ha?: ???0 if and only if ?0 lies outside the (1??)-level confidence interval for ?. Illustrate the preceding relationship by obtaining the appropriate? one-mean z-interval for the data below.

Suppose the mean height of women age 20 years or older in a certain country is 62.9 inches. One hundred randomly selected women in a certain city had a mean height of 61.6 inches. At the 1?% significance? level, the data provide sufficient evidence to conclude that the mean height of women in the city differs from the national mean. Assume that the population standard deviation of the heights of women in the city is 4.1 inches.

The confidence interval is ?(60.54?, 62.66?).

?(Round to two decimal places as? needed.)

Answer 1

Here, the null and the alternate hypothesis given are:

$\LARGE H_0: \mu = 62.9$

$\LARGE H_a: \mu \neq 62.9$

The test statistic here is computed as:

$\LARGE z^* = \frac{\bar X - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{62.9 - 61.6}{\frac{4.1}{\sqrt{100}}} =3.17$

Now as this is a two tailed test, we get the p-value from the standard normal tables as:

p = 2P( Z > 3.17 ) = 2*0.0008 = 0.0016

As the p-value here is 0.0016 < 0.01 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here are conclude that the data provide sufficient evidence to conclude that the mean height of women in the city differs from the national mean.

From standard normal tables, we get:

P( -2.576 < Z < 2.576 ) = 0.99

Therefore the confidence interval here is computed as:

$\LARGE \bar X \pm z^* \frac{\sigma}{\sqrt{n}}$

$\LARGE 61.6 \pm 2.576* \frac{4.1}{\sqrt{100}}$

$\LARGE 61.6 \pm 1.05616$

$\LARGE (60.54, 62.66)$

Which is the required confidence interval here for the population mean.