Question

In: Statistics and Probability

Within a school district, students were randomly assigned to one of two Math teachers - Mrs....

  1. Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students. At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15. Assume the variances are equal. At the significance level 0.10, can we conclude that Mrs. Jones is a more effective teacher than Mrs. Smith?

     a). Calculate the test statistic. (R code and R result).

     b). Find the p-value (R code and R result).

     c). Make your decision.

Solutions

Expert Solution

We want to test that Mrs. Jones is a more effective teacher than Mrs. Smith.

U1:- Mrs. Smith's students average test score

U2:-  Mrs. Jones' students average test score

We want to test Mrs. Jones is a more effective teacher than Mrs. Smith. ie. Mr.Jones have higher average scores.

The null & alternative Hypothesis:

Ho: U1 =U2

VS

Ha: U1<U2

a)

> # m1, m2: the sample means
> # s1, s2: the sample standard deviations
> # n1, n2: the same sizes
>
>
> n1<-30;n2<-25
> m1<-78;m2<-85
> s1<-10;s2<-15
> alpha=0.10
>
>
> # pooled standard deviation, scaled by the sample sizes
> se <- sqrt( ((1/n1) + (1/n2)) * (((n1-1)*s1^2) + ((n2-1)*s2^2))/(n1+n2-2) )
> df <- n1+n2-2
>
> # The test statistic:
>
> t <- (m1-m2)/se
>
> t
[1] -2.0656
>
>

b)
> # p-value
> pt(t,df)
[1] 0.02188548

c)

P-value = 0.02 < 0.10(level of significance)

So we reject Ho.

we may conclude that the data provide suficient evidence to connclude that the Mrs. Jones is a more effective teacher than Mrs. Smith.

***********R code ************

# m1, m2: the sample means
# s1, s2: the sample standard deviations
# n1, n2: the same sizes


n1<-30;n2<-25
m1<-78;m2<-85
s1<-10;s2<-15
alpha=0.10


# pooled standard deviation, scaled by the sample sizes
se <- sqrt( ((1/n1) + (1/n2)) * (((n1-1)*s1^2) + ((n2-1)*s2^2))/(n1+n2-2) )
df <- n1+n2-2

# The test statistic:

t <- (m1-m2)/se

t


# p-value
pt(t,df)

********* using a Function *********

# R code for left tailed test#

# m1, m2: the sample means
# s1, s2: the sample standard deviations
# n1, n2: the same sizes

# m0: the null value for the difference in means to be tested for. Default is 0.
# Alternative is
# equal.variance: whether or not to assume equal variance. Default is FALSE.
t.test2 <- function(m1,m2,s1,s2,n1,n2,m0=0,equal.variance=FALSE)
{
if( equal.variance==FALSE )
{
se <- sqrt( (s1^2/n1) + (s2^2/n2) )
# welch-satterthwaite df
df <- ( (s1^2/n1 + s2^2/n2)^2 )/( (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) )
} else
{
# pooled standard deviation, scaled by the sample sizes
se <- sqrt( (1/n1 + 1/n2) * ((n1-1)*s1^2 + (n2-1)*s2^2)/(n1+n2-2) )
df <- n1+n2-2
}
t <- (m1-m2-m0)/se
dat <- c(m1-m2, se, t, pt(t,df))
names(dat) <- c("Difference of means", "Std Error", "t", "p-value")
return(dat)
}

# you'll find this output agrees with that of t.test when you input x1,x2
t.test2( m1, m2, s1, s2, n1, n2,equal.variance=TRUE)


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