A random sample of 100 school administrators were conducted a Training Course. The sample mean and standard deviation were 90 and 5 respectively. In the standardization of the test, the mean was 85 and the standard deviation of 6. Test the significant difference using α=0.05 utilizing the p-value method. a. The p-value is 0.0002 0.0023 0.0046 0.0001
A random sample of 100 school administrators were conducted a Training Course. The sample
a. The p-value is
0.0002
0.0023
0.0046
0.0001
b. The Decision is
Accept the null hypothesis.There is a significant difference between the sample mean and the population mean.
Accept the null hypothesis.There is no significant difference between the sample mean and the population mean.
Reject the null hypothesis. There is no significant difference between the sample mean and the population mean.
Reject the null hypothesis. There is a significant difference between the sample mean and the population mean.
c. The null hypothesis is
μ<85
μ>85
μ=85
μ≤85
d. The alternative hypothesis is
μ≠85
μ<85
μ>85
μ=85
e. The value of the test statistic is
8.03
5.33
8.33
none of these
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