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Anastaziya [24]
2 years ago
12

A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec

ides 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?
Mathematics
2 answers:
pshichka [43]2 years ago
6 0

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, H_0 : \mu = 13 {means that the usual load is 13 credits}

Alternate Hypothesis, H_1 : \mu\neq 13 {means that the usual load is not 13 credits}

The test statistics we will use here is ;

              T.S. = \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, Xbar = sample mean = 13.65

               s = sample standard deviation = 1.91

               n = sample size = 100

So, test statistics = \frac{13.65 - 13}{\frac{1.91}{\sqrt{100} } } ~ t_9_9

                            = 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 = \frac{X-\mu}{\sigma} ~ N(0,1)

So, probability that student at this college takes 16 or more credits =     P(X \geq 16)

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

Ket [755]2 years ago
3 0

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

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

(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

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

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, <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

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

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Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =73,\sigma =10.5)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

We want to find the value of n that satisfy this condition:

P(71.5 < \bar X

And we can use the z score formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we have this:

P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z

And we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94

And by properties of the normal distribution we can express this like this:

P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z

If we solve for P(Z we got:

P(Z

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: z=-1.881

And using this we have this equality:

-1.881 = -0.14286 \sqrt{n}

If we solve for \sqrt{n} we got:

\sqrt{n} = \frac{-1.881}{-0.14286}=13.167

And then n=13.167^2 =173.369 and if we round up to the nearest integer we got n =174

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