In: Statistics and Probability

- 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.

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)

A school district developed an after-school math tutoring
program for high school students. To assess the effectiveness of
the program, struggling students were randomly selected into
treatment and control groups. A pre-test was given to both groups
before the start of the program. A post-test assessing the same
skills was given after the end of the program. The study team
determined the effectiveness of the program by comparing the
average change in pre- and post-test scores between the two groups....

12.16. Randomly selected groups of 120 parents and
150 teachers from one school district are surveyed about their
attitudes toward inclusion. One of the questions asks them whether
they oppose or support inclusions and their responses to this
question are recorded in the following table. The data were
analyzed using a chi square test. The obtained chi square value is
5.65, significant at the .02 level (p=.02).
GROUP
SUPPORT
OPPOSE
Parent
75
45
Teachers
72
78
a. Which
chi square test should...

Students in a biology class have been randomly assigned to one
of the two mentors for the laboratory portion of the class. A
random sample of final examination scores has been selected from
students supervised by each mentor, with the following
results:
Mentor A: 78, 78, 71, 89, 80, 93, 73, 76
Mentor B: 74, 81, 65, 73, 80, 63, 71, 64, 50, 80
At the 0.05 significance level, is there a difference in the mean
scores?

In one school district, there are 89 elementary school (K-5) teachers, of which 18 are male (or male-identifying). In a neighboring school district, there are 102 elementary teachers, of which 17 are male. A policy researcher would like to calculate the 99% confidence interval for the difference in proportions of male teachers.To keep the signs consistent for this problem, we will calculate all differences as p1−p2. That is, start with the percentage from the first school district and then subtract...

Fremont High School has 2100 students. One of the
statistics teachers at the school is interested in whether an
intervention program based on self-management improves attendance.
They randomly choose 80 students and randomly assign half of them
to either an experimental condition (self- management class) or a
control condition (distractor class on popular culture). At the end
of the semester, they measure the number of days missed for each
student. The teacher expects that the students in the
self-management class...

Students in an introductory economics course were assigned to
practical classes taught by various assistant teachers. The 21
students in the class of one of the assistant teachers obtained an
average score of 59.6 in the final exam and a standard deviation of
5.0. The 18 of the second obtained an average score in the final
exam of 85.2 and a standard deviation of 13.1. Suppose these data
can be considered independent random samples from populations that
follow a normal...

Conduct a one way between subjects ANOVA using SPSS
Students were randomly assigned to attend an ANOVA lecture that
contains one of the following approaches to instructions. 1)
Mathematical, 2) Conceptual, 3) Conceptual and Mathematical. After
the lecture, each student took an ANOVA exam. The table below
displays the scores by condition for each student. Are there any
significant differences between the groups?
Mathematical
Conceptual
Conceptual and Mathematical
75
80
95
81
92
97
93
91
87
90
73
96...

Salaries for teachers in a particular elementary school district
are normally distributed with a mean of $44,000 and a
standard deviation of $6,500. We randomly survey ten teachers from
that district.
1.Find the probability that the teachers earn a total of over
$400,000
2.If we surveyed 70 teachers instead of ten, graphically, how
would that change the distribution in part d?
3.If each of the 70 teachers received a $3,000 raise,
graphically, how would that change the distribution in part...

Salaries for teachers in a particular elementary school district
are normally distributed with a mean of $44,000 and a standard
deviation of $6,500. We randomly survey ten teachers from that
district. Find the 85th percentile for the sum of the sampled
teacher's salaries to 2 decimal places.

Salaries for teachers in a particular elementary school district
are normally distributed with a mean of $41,000 and a standard
deviation of $6,100. We randomly survey ten teachers from that
district.
A. Give the distribution of ΣX. (Round your answers to
two decimal places.)
ΣX - N ( , )
B. Find the probability that the teachers earn a total
of over $400,000. (Round your answer to four decimal places.)
C. Find the 80th percentile for an individual
teacher's salary....

ADVERTISEMENT

ADVERTISEMENT

Latest Questions

- Consider the initial value problem: y0 = 3 + x−y, y(0) = 1 (a) Solve it...
- Over the past year, M. D. Ryngaert & Co. had an increase in its current ratio...
- True or false: The parties in a jury trial have an unlimited number of peremptory challenges...
- The son, Seif, is born with sickle cell anemia. This means both parents carry the allele...
- A rapidly growing small firm does not have access to sufficient external financing to accommodate its...
- Which of the following is not an application of BI in Insurance ? Select one: a....
- A study compared three display panels used by air traffic controllers. Each display panel was tested...

ADVERTISEMENT