Mathematics, 17.03.2020 05:40 VamPL

A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor decides to randomly sample 100 students by using the registrar's database of students. The histogram below shows the distribution of the number of credits taken by these students. Sample statistics for this distribution are also provided. Min Q1 Median Mean SD Q3 Max

8 13 14 13.65 1.91 15 18

(a) Based on this data, would you accept or reject the hypothesis that the usual load is 13 credits?

(b) How unlikely is it that a student at this college takes 16 or more credits?

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Answer from: makaylacroy02

(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:

$Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18$

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,

$\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191$

(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:

$CI=\bar x\pm z_{\alpha/2}\times SE$

For 5% level of significance the two tailed critical value of z is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct the 95% confidence interval as follows:

$CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)$

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:

$P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z1.23)\\=1-P(Z$

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

Answer from: Quest

b. the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

Answer from: Quest

its 4 because 2 times 2 is 4, if you double the number two you get four