3. (32 Points) In order to assess how third-graders at a particular school are d

Question

3. (32 Points) In order to assess how third-graders at a particular school are doing in mathematics, you sample 50 students at random and look at their scores in mathematics on a standardized test. The school is doing well if the mean score exceeds 162. In these 50 students, you observe an average score of 168 with a variance of 10 points (a) (6 Points) Write a statistical model to describe the behavior of the scores (b) (8 Points) Construct a 95% confidence interval for the average mathematics score and give an interpretation of your interval. Specify a significance level and report your p-value sure to include comments in your code (c) (10 Points) Determine whether there is evidence to suggest that the students are doing well. (d) (8 Points-3 for b, 5 for c) Write an R program that carries out parts (b) and (c) for you. Be

Explanation / Answer

Part a

We are given

n = 50,

µ = 162,

= unknown

Xbar = 168,

Sample variance = 10,

Sample SD = sqrt(10) = 3.162278

We know that scores in mathematics on a standardized test follows a normal distribution. So, statistical model for variable score X is given as below:

X ~ N(µ = 162, = unknown)

Part b

We have to find 95% confidence interval for population mean.

We are given

n = 50,

Xbar = 168,

Sample variance = 10,

Sample SD = sqrt(10) = 3.162278

Confidence level = 95% = 0.95

= 0.05

/2 = 0.025

Degrees of freedom = n – 1 = 50 – 1 = 49

Critical value = t = 2.0096 (by using t-table)

Confidence interval = Xbar -/+ t*SD/sqrt(n)

Confidence interval = 168 -/+ 2.0096*3.162278/sqrt(50)

Confidence interval = 168 -/+ 0.8987

Lower limit = 168 - 0.8987 = 167.1013

Upper limit = 168 + 0.8987 = 168.8987

Confidence interval = (167.1013, 168.8987)

We are 95% confident that the average population score will lies between 167.1013 and 168.8987.

Part c

Here, we have to use one sample t test for the population mean to check the claim or hypothesis whether there is any evidence to suggest that the students are doing well.

Students are doing well if Xbar > 162.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: µ = 162

Alternative hypothesis: Ha: µ > 162

This is a one tailed test. This is a right tailed or upper tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[SD/sqrt(n)]

We are given

n = 50,

µ = 162,

= unknown

Xbar = 168,

Sample variance = 10,

Sample SD = sqrt(10) = 3.162278

Degrees of freedom = 50 – 1 = 49

Significance level = = 0.05

t = (168 – 162) / [3.162278/sqrt(50)]

t = 13.4164

P-value = 0.00

P-value < , so we reject the null hypothesis.

There is sufficient evidence that students are doing well.

Part d

The R program for part b is given as below:

> a = 168

> s = 3.162278

> n=50

> error = qt(0.975, df=n -1)*s/sqrt(n)

> left = a - error

> right = a + error

> left

[1] 167.1013

> right

[1] 168.8987

The R program for part c is given as below:

> n=50

> mu=162

> xbar=168

> SD=3.162278

> alpha=0.05

> df=n-1

> t=(xbar - mu)/(SD/sqrt(n))

> p = dt(t,df,log=FALSE)

> df

[1] 49

> t

[1] 13.41641

> p

[1] 7.206281e-18

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