Respuesta :
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 (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.
The information provided is:
[tex]Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18[/tex]
The sample size is, n = 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,
[tex]\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191[/tex]
(a)
The null hypothesis is:
H₀: The usual load is 13 credits, i.e. μ = 13.
Assume that the significance level of the test is, α = 0.05.
Construct a (1 - α) % confidence interval for population mean to check the claim.
The (1 - α) % confidence interval for population mean is given by:
[tex]CI=\bar x\pm z_{\alpha/2}\times SE[/tex]
For 5% level of significance the two tailed critical value of z is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Construct the 95% confidence interval as follows:
[tex]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)[/tex]
As the null value, μ = 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:
[tex]P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z<1.23)\\=1-0.8907\\=0.1093[/tex]
Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.
Answer:
(a) We reject the null hypothesis that usual load is 13 credits.
(b) Probability that student at this college takes 16 or more credits = 0.10935
Step-by-step explanation:
We are given that the histogram below shows the distribution of the number of credits taken by these students;
Min Q1 Median Mean SD Q3 Max
8 13 14 13.65 1.91 15 18
Also, the counselor decides to randomly sample 100 students by using the registrar's database of students, i.e., n = 100.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 13 {means that the usual load is 13 credits}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 13 {means that the usual load is not 13 credits}
The test statistics we will use here is ;
T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, Xbar = sample mean = 13.65
s = sample standard deviation = 1.91
n = sample size = 100
So, test statistics = [tex]\frac{13.65 - 13}{\frac{1.91}{\sqrt{100} } }[/tex] ~ [tex]t_9_9[/tex]
= 3.403
Now, since significance level is not given to us so we assume it to be 5%.
At 5% significance level, the t tables gives critical value of 1.987 at 99 degree of freedom. Since our test statistics is more than the critical value which means our test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that the usual load is not 13 credits.
(b) Let X = credits of students
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So, probability that student at this college takes 16 or more credits = P(X [tex]\geq[/tex] 16)
P(X [tex]\geq[/tex] 16) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{16-13.65}{1.91}[/tex] ) = P(Z [tex]\geq[/tex] 1.23) = 1 - P(Z < 1.23)
= 1 - 0.89065 = 0.10935 or 11%
Therefore, probability that student at this college takes 16 or more credits is 0.10935.
