Math / Data & Statistics

Question1
In Country A, the population mean height for 3-year-old boys is 39 inches. Suppose a random sample of 153 -year-old boys from Country B showed a sample mean of 38.7 inches with a standard deviation of 4 inches. The boys were independently sampled. Assume that heights are Normally distributed in the population. Complete parts a through c below. a. Determine whether the population mean for Country B boys is significantly different from the Country A mean. Use a significance level of 0.05 .
Which of the following correctly states $\mathrm{H}_{0}$ and $\mathrm{H}_{a}$ ? A. $\begin{array}{l}H_{0}: \mu=39 \\ H_{0}: \mu \neq 39\end{array}$ $H_{0}: \mu \geq 39$ c. $H_{0}: \mu=39$ B. $H_{a}: \mu<39$ . $H_{a}: \mu<39$
$H_{0}: \mu>39$ D. $H_{a}: \mu \leq 39$ $H_{0}: \mu \neq 39$ F. $\begin{array}{l}H_{0}: \mu=39 \\ H_{a}: \mu>39\end{array}$

Studdy Solution

STEP 1

1. The population mean height for 3-year-old boys in Country A is 39 inches.
2. A random sample of 153 3-year-old boys from Country B has a sample mean height of 38.7 inches and a standard deviation of 4 inches.
3. Heights are Normally distributed in the population.
4. We are using a significance level of 0.05.

STEP 2

1. Define the null and alternative hypotheses.
2. Calculate the test statistic.
3. Determine the critical value or p-value.
4. Make a decision to accept or reject the null hypothesis.

STEP 3

Define the null and alternative hypotheses. We want to test if the mean height of boys in Country B is significantly different from that in Country A.
The correct hypotheses are: - Null hypothesis ( $H_0$ ): $\mu = 39$ - Alternative hypothesis ( $H_a$ ): $\mu \neq 39$
This corresponds to option A.

STEP 4

Calculate the test statistic using the formula for the one-sample t-test:
$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$
where: - $\bar{x} = 38.7$ (sample mean) - $\mu_0 = 39$ (population mean under the null hypothesis) - $s = 4$ (sample standard deviation) - $n = 153$ (sample size)
Substitute the values:
$t = \frac{38.7 - 39}{\frac{4}{\sqrt{153}}}$
Calculate the denominator:
$\frac{4}{\sqrt{153}} \approx \frac{4}{12.369} \approx 0.323$
Calculate the test statistic:
$t = \frac{-0.3}{0.323} \approx -0.929$

STEP 5

Determine the critical value or p-value for a two-tailed test at the 0.05 significance level with $n - 1 = 152$ degrees of freedom.
Using a t-distribution table or calculator, find the critical t-value for $df = 152$ at $\alpha = 0.05$ for a two-tailed test. The critical t-value is approximately $\pm 1.976$ .

STEP 6

Make a decision to accept or reject the null hypothesis.
Since the calculated t-value $-0.929$ is within the range $-1.976$ to $1.976$ , we fail to reject the null hypothesis.
Conclusion: There is not enough evidence to suggest that the population mean height for Country B boys is significantly different from the Country A mean at the 0.05 significance level.

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