Question

Question 11

Dr. Calvin Broadus wants to determine whether marijuana reduces anxiety. He draws a sample of n = 16 individuals. The population from which they were selected exhibit a score of µ = 30 on an anxiety measure. After “treatment” is administered to the individuals, the sample mean is found to be M = 36 and the sample variance is s² = 48. If the researcher decides to test at the α = .05 level, what are degrees of freedom and tcritical value? df= 15, critical t = -1.74 df= 16, critical t = +/-1.74 df= 15, critical t = -2.11 df= 16, critical t = +/-2.11

QUESTION 12

Continuing Scenario A (from Question #10): Dr. Calvin Broadus wants to determine whether marijuana reduces anxiety. He draws a sample of n = 16 individuals. The population from which they were selected exhibit a score of µ = 30 on an anxiety measure. After “treatment” is administered to the individuals, the sample mean is found to be M = 36 and the sample variance is s2 = 48. In determining whether the data are sufficient to conclude if marijuana reduces anxiety, what is the observed test statistic value?

3.46

-2.17

.07

2.23

QUESTION 13

Continuing Scenario A (from Question #10), based on the observed test statistics, would Dr. Broadus reject or retain the null hypothesis?

Reject Null Hypothesis (H0)

Retain Null Hypothesis (H0)

QUESTION 14

Regardless of your response for the previous question, consider a situation in which Dr. Broadus decided to reject the null hypothesis. If he rejected the null hypothesis, what would Dr. Broadus conclude about the effect of marijuana on anxiety?

Marijuana significantly reduces anxiety.

Marijuana significantly increases anxiety.

There is insufficient evidence to determine whether marijuana influences anxiety.

Marijuana significantly influences anxiety.

QUESTION 15

Continuing Scenario A (from Question #10): Dr. Calvin Broadus wants to determine whether marijuana reduces anxiety. He draws a sample of n = 16 individuals. The population from which they were selected exhibit a score of µ = 30 on an anxiety measure. After “treatment” is administered to the individuals, the sample mean is found to be M = 36 and the sample variance is s2 = 48. Construct a 95% confidence interval for an estimate of the population mean.

29.63 to 33.52

28.74 to 37.26

32.35 to 39.65

30.21 to 37.79

QUESTION 16

Dr. Gupta conducts a study to examine whether eating cookies influences memory. She draws a sample of n = 36 participants. The population from which they were selected had a mean memory score of µ = 45 on a memory test. After eating cookies, the sample mean is found to be M = 58 and the estimated standard error is 0.74. Calculate the effect size in terms of the proportion of variance accounted for (r2).

0.93

0.73

0.23

0.55

QUESTION 17

Research Scenario B: a teacher gives a reading skills test to a third-grade class of n = 25 at the beginning of the school year. To evaluate whether students' reading performance changed over the course of the year, they are tested again at the end of the year. Their test scores showed an average improvement of MD = 4.8 points with s2 = 75. Is this a one-tailed or two-tailed test?

Two-tailed test

One-tailed test

QUESTION 18

Continuing with Scenario B (from Question #17), if the teacher decides to test at the alpha = .01 level, what are the degrees of freedom and critical t value?

df = 27, critical t = 2.47

df = 24, critical t = 2.80

df = 48, critical t = 2.70

df = 25, critical t = +/-2.47 3 points

QUESTION 19

Continuing with Research Scenario B (from Question #17): a teacher gives a reading skills test to a third-grade class of n = 25 at the beginning of the school year. To evaluate whether students' reading performance changed over the course of the year, they are tested again at the end of the year. Their test scores showed an average improvement of MD = 4.8 points with s2 = 75. What is the observed test statistic value? Be sure to round to two decimal places for the final answer. 6 points

QUESTION 20

Continuing with Scenario B, given the observed t-statistic, should the teacher reject or retain the null hypothesis (H0)?

Retain Null Hypothesis (H0)

Reject Null Hypothesis (H0)

Answer 1

Question 11

Dr. Calvin Broadus wants to determine whether marijuana reduces anxiety.

From this claim the test is left tail test.

So we have negative critical t value

critical t = "=-TINV(0.1,15)" = -1.753

Degrees of freedom = n - 1 = 16 - 1 = 15

So correct choice is first option "df= 15, critical t = -1.74"

QUESTION 12

The observed test statistic value is as follow

QUESTION 13

Continuing Scenario A (from Question #10), based on the observed test statistics, would Dr. Broadus reject or retain the null hypothesis?

Retain Null Hypothesis (H0)

Because critical value = -1.74 < t = 3.46

Note that here null hypothesis is marijuana does not reduces anxiety.

QUESTION 14

Regardless of your response for the previous question, consider a situation in which Dr. Broadus decided to reject the null hypothesis. If he rejected the null hypothesis, what would Dr. Broadus conclude about the effect of marijuana on anxiety?

If he rejected the null hypothesis, Then he may conclude about the effect of marijuana on anxiety as

Marijuana significantly increases anxiety.

Because critical value for right tail test is 1.74 < t = 3.46.

QUESTION 15

Let's write the given information

n = sample size = 16

= sample mean = 36

s = sample standard deviation = 48

Let's use minitab

The command is Stat>>>Basic Statistics >>1 sample t...

Select summary Statistics

then click on Option select level of confidence = 95

Alternative " not equal"

Look the following image

then click on Ok again click on OK

So we get the following output

From the above output the 95% confidence interval for an estimate of the population mean is as follow:

32.31 to 39.69

So the correct option is 32.35 to 39.65 (because it is closed to 32.31 to 39.69 ).

p- value = 0.005

Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.005 < 0.01 so we used first rule.

That is we reject null hypothesis and accept alternative hypothesis.