Solution :
=
103
= 107.8
S = 13.7
n = 42
This is the two tailed test .
The null and alternative hypothesis is ,
H0 :
= 103
Ha :
> 103
Test statistic = T
= (
-
) / S /
n
= (107.8-103) /13.7 /
42
= 2.27
P(z > 2.27) = 1 - P(z < 2.27) = 0.0142
P-value = 0.0142
= 0.05
p = 0.0142 < 0.05, it is concluded that the null hypothesis is rejected..
There is enough evidence to claim that the population mean μ is greater than 103, at the 0.05 significance level.
Score: 0 of 1 pt 7 of 11 (9 complete) X 5.4.17 The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual. For a sample of n-75, find the probability of a sample mean being greater than 214 if213 and 6.2. For a sample of n = 75, the probability of a sample mean being greater than 214 213 and a = 6.2 is 0808 (Round...
2 of 11 (0 complete) TO Score: 0 of 1 pt 10.4.3-T Is Question Help A simple random sample of size n= 15 is drawn from a population that is normally distributed. The sample mean is found to be x = 19.2 and the sample standard deviation is found to be s=6.3. Determine if the population mean is different from 25 at the a=0.01 level of significance. Complete parts (a) through (d) below (a) Determine the null and alternative hypotheses....