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:
![Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18](https://tex.z-dn.net/?f=Min.%3D8%5C%5CQ_%7B1%7D%3D13%5C%5CMedian%3D14%5C%5CMean%3D13.65%5C%5CSD%3D1.91%5C%5CQ_%7B3%7D%3D15%5C%5CMax.%3D18)
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,
![\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%7D%3D%5Cbar%20x%3D13.65%5C%5CSE%3D%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B1.91%7D%7B%5Csqrt%7B100%7D%7D%3D0.191)
(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:
![CI=\bar x\pm z_{\alpha/2}\times SE](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ctimes%20SE)
For 5% level of significance the two tailed critical value of <em>z</em> is:
![z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.05%2F2%7D%3Dz_%7B0.025%7D%3D1.96)
Construct the 95% confidence interval as follows:
![CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ctimes%20SE%5C%5C%3D13.65%5Cpm%20%281.96%5Ctimes0.191%29%5C%5C%3D13.65%5Cpm0.3744%5C%5C%3D%2813.2756%2C%2014.0244%29%5C%5C%3D%2813.28%2C%2014.02%29)
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:
![P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z](https://tex.z-dn.net/?f=P%28X%5Cgeq%2016%29%3DP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5Cgeq%20%5Cfrac%7B16-13.65%7D%7B1.91%7D%29%5C%5C%3DP%28Z%3E1.23%29%5C%5C%3D1-P%28Z%3C1.23%29%5C%5C%3D1-0.8907%5C%5C%3D0.1093)
Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.