Question

A random sample of 49 measurements from a population with population standard deviation o 3 had a sample mean of x, 9. An independeent random sample of sample mean of x, 11. Test the claim that the population means are 64 measurements from a second population with population standard deviation a2 4 had different. Use level of significance 0.01. (a) What distribution does the sample test statistic follow? Explain. The student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal distribution. Samples are independent, the population standard deviations are known, and the sample sizes are sufficiently large. OThe standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The student's t. We assume that both population distributions are approximately normal with known standard deviations. (b) State the hypotheses. Ho: H 2i H1: 2 Ho: 2: H:1 2 :H = 'ri :°H O OHo: z; Hy: 1< H2 (c) Compute x1-x2. 1-X2-2 Compute the corresponding sample distribution value. (Test the difference u, -u. Round your answer to two decimal places.) -0.25 (d) Find the P-value of the sample test statistic. (Round your answer to four decimal places.)

Answer 1

(a) Since both sample sizes are greater than 30, we can comfortably use the normal distribution.

(b) The null and alternative hypothesis are given by:

$\dpi{80} \fn_phv \small H_0=\mu_1-\mu_2=0 \rightarrow H_0=\mu_1=\mu_2, H_1=\mu_1\neq \mu_2$

(c) Difference between the two given sample means is -2.

To calculate the sample distribution value, we need to find the sample standard deviation. It can be calculated as follows:

$\dpi{80} \fn_phv \small \sigma=\sqrt{\frac{3^2}{49}&plus;\frac{4^2}{64}}=\sqrt{0.1836&plus;0.25}=0.6584$

Calculating the sample distribution value:

$\dpi{80} \fn_phv \small z=\frac{-2-0}{0.6584}=-3.0376$

(d) The p-value for the calculated test statistic can be found using the z table. It is 0.0012

(e) We compare the calculated test statistic to the test statistic in the z table for a significance level of 0.01. We look for the value corresponding to a significance level of 0.005 because this is a two tailed test. According to the z table this value is -2.58. Since the calculated z value is less than this value, we reject the null hypothesis at a significance level of 0.01 and conclude that the data are statistically significant.

(f) We reject the null hypothesis, there is sufficient evidence that there is a difference between the population means.

(g) To find the 99% confidence interval around the difference of means:

$\dpi{80} \fn_phv \small Pr(-2.58<\frac{-2-(\mu_1-\mu_2)}{0.6584}<2.58)=0.99 \newline \rightarrow Pr(-0.3013<\mu_1-\mu_2<0.3013)=0.99$

Lower limit of the confidence interval is -0.3013

Upper limit of the confidence interval is 0.3013.