Question

Q1. Consider the population of adult female residents in Melbourne (or Jupiter if you prefer). Our focus is on the population mean height. Let height be called X. Assume we do not know (population standard deviation) or the population mean, u. We take a sample of adult female residents in Melbourne (n=100) and calculate the sample mean height as 70 cm and the sample standard deviation of 25. i) Test the null hypothesis that u=120, against the alternative that u#120. Use a 5% level of significance. Interpret your results. (5 marks) ii) Now use a one sided test 5% significance level Ho=2120, H1 = u <120. Interpret your results. (5 marks)

Please help.

Answer 1

i)

JO = 120 n = 100 S=25 X = 70 a= 0,05

The test hypothesis is

Null Hypothesis > Horl=MO Alternate Hypothesis > Hippo

This is a two-sided test because the alternative hypothesis is formulated to detect differences from the hypothesized mean value of 30 on either side.
Now, the value of test static can be found out by following formula:
$\\t_0 = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \\t_0 = \frac{70.0- 120.0}{25.0/\sqrt{100}} \\t_0 = -20.0$
$\\\\\text{For }\alpha = 0.05, t_{\alpha/2}=t_{0.025}=1.9842.\\\\\text{ Since }t_0 = -20.0 < -1.9842 = -t_{0.025},\text{ we reject the null hypothesis }H_0:\mu=120.0\text{ in favor of the alternative hypothesis }H_1:\mu\ne120.0\text{ at }\alpha=0.05.$

ii)
The test hypothesis is

Null Hypothesis > Ho :μ2 μο Alternate Hypothesis --> H1 :μ< μο.

Now, the value of test static can be found out by following formula:
$\\t_0 = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \\t_0 = \frac{70.0- 120.0}{25.0/\sqrt{100}} \\t_0 = -20.0$
Since the sample size is n = 100, degrees of freedom on the t-test statistic are n-1 = 100-1 = 99
This implies that
$\\t_{\alpha, n-1} = t_{0.05, 99} = 1.6604 \text{Since, the t distribution is symmetric about zero, so }-t_{0.05,99} \\\\\text{Since }t_0 = -20.0<-1.6604=-t_{0.05},\text{ we reject the null hypothesis }H_0:\mu=120.0\text{ in favor of the alternative hypothesis }H_1:\mu >120.0\text{ at }\alpha = 0.05.$
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