Question

(40 POINTS!!!) A scientist claims that pneumonia causes weight loss in mice. The table shows the weights​ (in grams) of six mice before infection and two days after infection. At a=0.10​, is there enough evidence to support the​ scientist's claim? Assume the samples are random and​ dependent, and the population is normally distributed. Complete parts​ 1 through 4 below.

Mouse Weight_(before) Weight_(after)
1 21.7 21.6
2 21.6 21.7
3 22.6 22.4
4 19.5 19.6
5 22.2 22.2
6 23.6 23.6

Part 1)
Identify the claim and state H0 and Ha .
What is the​ claim?
A. Weight gain causes pneumonia in mice.
B. Pneumonia causes weight gain in mice.
C. Pneumonia causes weight loss in mice.
D. Weight loss causes pneumonia in mice.

Let ud be the hypothesized mean of the difference in the weights ​(before​after). What are H0 and Ha​?

Part 2)
Find the critical​ value(s) and identify the rejection​ region(s).
Select the correct choice below and fill in any answer boxes to complete your choice.
​(Round to three decimal places as​ needed.)
A. t> ?
B. t < ? or t > ?
C. t < ?

Part 3)
Calculate d- and sd.
d-=?
sd=?
Find the standardized test statistic t.
t=?

Part 4)
Decide whether to reject or fail to reject the null hypothesis and interpret the decision in the context of the original claim.
▼
Reject
Fail to reject
the null hypothesis. There
▼
is
is not
enough evidence to
▼
reject
support
the claim that
▼
weight gain
weight loss
pnuemonia
causes
▼
pnuemonia
weight loss
weight gain
in mice.

Answer 1

Explanation:

Part 1:

The claim is that pneumonia causes weight loss in mice.

H0 (null hypothesis): The mean difference in weights (before and after) is equal to zero.

Ha (alternative hypothesis): The mean difference in weights (before and after) is less than zero.

Answer: C. Pneumonia causes weight loss in mice.

Part 2:

Since the alternative hypothesis is that the mean difference is less than zero (weight loss), the rejection region is on the left side of the t-distribution.

We will perform a one-tailed test.

The critical value for a one-tailed test with a significance level of 0.10 and 5 degrees of freedom is approximately -1.833.

Answer: B. t < -1.833

Part 3:

To calculate d- (the mean difference) and sd (standard deviation of the differences):

d- = Σ (Weight_after - Weight_before) / n

= (21.6 - 21.7) + (21.7 - 21.6) + (22.4 - 22.6) + (19.6 - 19.5) + (22.2 - 22.2) + (23.6 - 23.6) / 6

= -0.033

sd = √[ Σ (Weight_after - Weight_before - d-)² / (n-1) ]

= √[ (-0.033 - (-0.033))² + (-0.033 - (-0.033))² + (-0.033 - (-0.033))² + (-0.033 - (-0.033))² + (-0.033 - (-0.033))² + (-0.033 - (-0.033))² / (6-1) ]

= 0

Note: The standard deviation (sd) is zero because all the differences are the same.

The standardized test statistic t is calculated as:

t = d- / (sd / √n)

= -0.033 / (0 / √6)

= undefined

Answer: d- = -0.033, sd = 0, t = undefined

Part 4:

Since the t-value is undefined (division by zero), we cannot calculate the test statistic. Therefore, we cannot make a decision to reject or fail to reject the null hypothesis.

The interpretation is that there is not enough evidence to support or reject the claim that pneumonia causes weight loss in mice based on the given data.