A scientist claims that pneumonia causes weight loss in mice. The table attached here shows the weights(in grams) of six mice before infection and two days after infection. At α = 0.01, is there enough evidence to support the scientist’s claim? Assume the samples are random anddependent, and the population is normally distributed.MouseWeight (Before)Weight (After)121.521.3223.823.8322.922.7420.820.7521.321.2621.721.7Calculate dbar (sample mean of the differences)and Sd(sample standard deviation of the differences) and decide whether to reject or fail to reject the null. Interpret the results in context of the claim and show all work.

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Description: A scientist claims that pneumonia causes weight loss in mice. The table attached here shows the weights(in grams) of six mice before infection and two days after infection. At α = 0.01, is there enough evidence to support the scientist’s claim? Assu

Given Information : A scientist claims that pneumonia causes weight loss in mice. The provided shows the weights(in grams) of six mice before infection and two days after infection. The following table is obtained: Wt. (Before) Wt. (After) Difference = Wt. (Before)- Wt. (After) 21.5 21.3 0.2 23.8 23.8 0 22.9 22.7 0.2 20.8 20.7 0.1 21.3 21.2 0.1 21.7 21.7 0 Average 22 21.9 0.1 St. Dev. 1.124 1.147 0.089 n 6 6 6 From the sample data, it is found that the corresponding sample means are: \bar X_1 = 22X¯1=22 , X¯2 =21.9 Also, the provided sample standard deviations are: s_1 = 1.124s1=1.124s_2 = 1.147 , s2=1.147 and the sample size is n = 6. For the score differences we have \bar D = 0.1 D¯ =0.1 , SD=0.089 A t-test for two paired samples is a hypothesis test that attempts to make a claim about the population means (μ1 and μ2). More specifically, a t-test uses sample information to assess how plausible it is for difference \mu_1μ1 - \mu_2μ2 to be equal to zero. Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μD = 0 Ha: μD > 0 This corresponds to a right-tailed test, for which a t-test for two paired samples be used. Rejection Region Based on the information provided, the significance level is \alpha = 0.01α=0.01, and the degrees of freedom are df = 5df=5. Hence, it is found that the critical value for this right-tailed test is t_c = 3.365tc=3.365, for \alpha = 0.01α=0.01 and df = 5df=5. The rejection region for this right-tailed test is R = {t: t > 3.365}R=t:t>3.365. Test Statistics The t-statistic is computed as shown in the following formula: Decision about the null hypothesis Since it is observed that t = 2.739 \le t_c = 3.365t=2.739 ≤ tc=3.365, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.0204, and since p=0.0204≥0.01, it is concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean \mu_1μ1 is greater than \mu_2μ2, at the 0.01 significance level. That is , There is not enough evidence to claim that the pneumonia causes weight loss in mice .

Answered: A scientist claims that pneumonia…

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A scientist claims that pneumonia causes weight loss in mice. The table attached here shows the weights(in grams) of six mice before infection and two days after infection. At α = 0.01, is there enough evidence to support the scientist’s claim? Assume the samples are random anddependent, and the population is normally distributed.

Mouse	Weight (Before)	Weight (After)
1	21.5	21.3
2	23.8	23.8
3	22.9	22.7
4	20.8	20.7
5	21.3	21.2
6	21.7	21.7

Calculate dbar (sample mean of the differences)and Sd(sample standard deviation of the differences) and decide whether to reject or fail to reject the null. Interpret the results in context of the claim and show all work.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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