Birth Weights of Male Babies| Confidence Interval and Hypothesis Testing for Population Mean and Standard Deviation| Statistics

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 \bar{x}=35.3,\:and\:s=7.2.

Because we want a 90% confidence interval, we have \alpha =1-0.90=0.10. The sample size is 200. So, we have to find t_{\alpha /2}=t_{0.05} with 200-1=199 degrees of freedom. Using technology, the value of t_{0.05}=1.65.

The lower bound is

\bar{x}-t_{\alpha /2}\cdot \frac{s}{\sqrt{n}}

=35.3-1.65\cdot \frac{7.2}{\sqrt{200}}

=34.5\:hg

The upper bound is

\bar{x}+t_{\alpha /2}\cdot \frac{s}{\sqrt{n}}

=35.3+1.65\cdot \frac{7.2}{\sqrt{200}}

=36.1\:hg

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

H_0:\:\mu =37.0

H_1:\:\mu \ne 37.0

The test statistic is

t_0=\frac{\bar{x}-\mu _0}{s/\sqrt{n}}=\frac{35.3-37}{7.2/\sqrt{200}}=-3.34

Since this is a two-tailed test, we determine the critical values at the \alpha \:/2=0.05 level of significance with 200-1=199 degrees of freedom. From PART A, the critical values are t_{0.05}=\pm 1.65. The rejection regions are t<-1.65,\:or\:t>1.65 as shown below.

0.05-two-tailed-t

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

H_0:\sigma =6.75

H_1:\sigma \ne 6.75

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

The test statistic is

\chi _0^2=\frac{\left(n-1\right)s^2}{\sigma _0^2}

\chi _0^2=\frac{\left(200-1\right)\left(7.2\right)^2}{\left(6.75\right)^2}

\chi _0^2=226.42

Because this is a two-tailed test, we determine the critical \chi ^2 values at the \alpha /2=0.025 level of significance with 199 degrees of freedom to be \chi ^2_L=161.8 and \chi ^2_R=240.0. The rejection regions are \chi ^2<161.8\:and\:\chi ^2>240.0

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. 

 

 

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