Standardized measures seem to indicate that the average level of anxiety has inc

Question

Standardized measures seem to indicate that the
average level of anxiety has increased gradually over
the past 50 years (Twenge, 2000). In the 1950s, the
average score on the Child Manifest Anxiety Scale
was u, = 15.1. A sample of n = 16 of today's children
produces a mean score of M = 23.3 with SS = 240.
a. Based on the sample, has there been a significant
change in the average level of anxiety since the
1950s? Use a two-tailed test with a = .01.
b. Make a 99% confidence interval estimate of today's
population mean level of anxiety.
c. Write a sentence that demonstrates how the
outcome of the hypothesis test and the confidence
interval would appear in a research report.

Explanation / Answer

Solution:

Part a

Here, we have to use one sample t test for population mean. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: There is no significant change in the average level of anxiety since the 1950s.

Alternative hypothesis: Ha: There is a significant change in the average level of anxiety since the 1950s.

H0: µ = 15.1 vs. Ha: µ ? 15.1

This is a two tailed test.

We are given

n = 16,

M = 23.3,

SS = 240,

? = 0.01,

df = n – 1 = 16 – 1 = 15

Standard Deviation = S = sqrt(SS/(n – 1))

S = sqrt(240/(16 – 1)) = sqrt(240/15) = sqrt(16) = 4

Test statistic = t = (M - µ) / [S/sqrt(n)]

t = (23.3 – 15.1) / [4/sqrt(16)] = 8.2/1 = 8.2

P-value = 0.00 (by using t-table or excel)

P-value < ? = 0.01

So, we reject the null hypothesis that there is no significant change in the average level of anxiety since the 1950s.

There is sufficient evidence to conclude that there is a significant change in the average level of anxiety since the 1950s.

Part b

Here, we have to construct 99% confidence interval for estimate of today’s population mean level of anxiety.

We are given

Xbar = 23.3, S = 4, n = 16, c = 0.99, df = 15

Critical t value = 2.9467

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

Confidence interval = 23.3 -/+ 2.9467*4/sqrt(16)

Confidence interval = 23.3 -/+ 2.9467*1

Confidence interval = 23.3 -/+ 2.9467

Lower limit = 23.3 – 2.9467 = 20.3533

Upper limit = 23.3 + 2.9467 = 26.2467

Confidence interval = (20.3533, 26.2467)

Part c

From hypothesis test in part a, at 1% level of significance there is sufficient evidence to conclude that there is a significant change in the average level of anxiety since the 1950s.We are 99% confidence that the population mean level of anxiety will be lies between (20.3533, 26.2467). The value µ = 15.1 is not lies within this confidence interval and therefore we reject the null hypothesis that there is no any significant change in the average level of anxiety since the 1950s.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.