### Birth weights of male babies at one hospital were recorded for 200 births of boys. A mean of that sample was 35.3 hg and a standard deviation of 7.2 hg.

a) Construct a 90% confidence interval for the mean of all boy baby weights.

b) A medical Journal claims that the mean weight of male babies is 37.0 hg. Test that claim at a 0.10 level of significance.

c) Use a 0.05 level of significance to test a claim that the real standard deviation of boy birth weights is 6.75 hg.

**SOLUTION:**

**PART A**

So, we are given .

Because we want a 90% confidence interval, we have . The sample size is 200. So, we have to find with degrees of freedom. Using technology, the value of .

The lower bound is

The upper bound is

We are 90% confident that the mean of all baby boy weights is between 34.5 hg and 36.1 hg.

**PART B**

The null and alternative hypotheses are

The test statistic is

Since this is a two-tailed test, we determine the critical values at the level of significance with degrees of freedom. From PART A, the critical values are . The rejection regions are as shown below.

Since the test statistic is in the rejection region, we REJECT the null hypothesis. There is insufficient evidence at 90% confidence level to support the claim that the mean weight of all baby boys is 37.0 hg.

**PART C**

The null and alternative hypotheses are

So, this is a two-tailed test, and the level of significance is 0.05.

The test statistic is

Because this is a two-tailed test, we determine the critical values at the level of significance with 199 degrees of freedom to be and . The rejection regions are

Since the test statistic is not in the rejection region, we FAIL TO REJECT the null hypothesis. There is sufficient evidence at 0.05 level of significance to support the claim that the real standard deviation is 6.75 hg.