(20 PTS) Please help!!!! I've been stuck for 1 hour already! :(
1 answer:
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Answer:
(a) The usual load is not 13 credits.
(b) The probability that a a student at this college takes 16 or more credits is 0.1093.
Step-by-step explanation:
According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.
The information provided is:
The sample size is, <em>n</em> = 100.
The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.
So,
(a)
The null hypothesis is:
<em>H</em>₀: The usual load is 13 credits, i.e. <em>μ</em> = 13.
Assume that the significance level of the test is, <em>α</em> = 0.05.
Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.
The (1 - <em>α</em>) % confidence interval for population mean is given by:
For 5% level of significance the two tailed critical value of <em>z</em> is:
Construct the 95% confidence interval as follows:
As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.
Thus, it can be concluded that the usual load is not 13 credits.
(b)
Compute the probability that a a student at this college takes 16 or more credits as follows:
Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.