Answer :
Answer:
1- A critical value right-tailed Normality test.
2- Yes, they did.
3-
Null hypothesis
[tex]\bf H_0:[/tex] The national average on the statistics achievement test is 60
Alternative hypothesis
[tex]\bf H_a:[/tex] The average on the experimental program for statistics education is greater than 60
4- 0.05 or 5%
5- A Type 1 Error is the error we make when we reject the null hypothesis given that it is true.
Step-by-step explanation:
1)
Since the sample size is greater than 30, the population follows a Normal distribution, and we suspect the average is greater than the established one, the appropriate test to use is a critical value right-tailed Normality test.
2)
To check if the students scored significantly above the national average in this special program to a significance level α = 0.05, we must check if the z-statistic given by the sample falls to the right of 1.64, since the area under the Normal N(0;1) to the right of 1.64 equals 0.05
Our z-statistic is given by the formula
[tex]\bf z=\frac{\bar x - \mu}{\sigma/\sqrt{n}}[/tex]
where
[tex]\bf \bar x[/tex] is the mean of the sample
[tex]\bf \mu[/tex] is the mean of the null hypothesis
[tex]\bf \sigma[/tex] is the standard deviation
n is the sample size
Our z-statistic is then
[tex]z=\frac{64.8-60}{12/\sqrt{36}}=2.4[/tex]
Since 2.4 > 1.64, the students did score significantly above the national average in this special program to a significance level α = 0.05
3)
Null hypothesis
[tex]\bf H_0:[/tex] The national average on the statistics achievement test is 60
Alternative hypothesis
[tex]\bf H_a:[/tex] The average on the experimental program for statistics education is greater than 60
4)
The probability of a Type 1 Error is 0.05 or 5% (the significance level)
5)
A Type 1 Error is the error we make when we reject the null hypothesis given that it is true.