Question

Use R. Provide Solution and R Code within each problem. For this section, use the mtcars dataset available in R (you do not need to download any packages). Assume that the distribution of weights (wt) are approximately normally distributed within transmission types (am) and the samples are drawn independently.

a.Write the null and alternative hypothesis to determine if the mean weight of cars differs by transmission type in the population.

b.Using = 0.05, conduct a statistical test for the hypotheses in 1. . Interpret your conclusion in the context of the problem.

i.Pvalue

ii.Conclusion

c.Calculate a 95% confidence interval for the difference in the mean weights of cars with automatic transmissions and standard transmissions.

d.Write the null and alternative hypothesis to determine if the mean weight of automatic cars is less than the mean weight of standard transmission cars in the population. e.Using = 0.05, conduct a statistical test for the hypotheses in 4. Interpret your conclusion in the context of the problem.

i.Pvalue

ii.Conclusion

Answer 1

a. Let $\mu$ 1 be the mean weight of cars where am =0 and $\mu$ 2 be the mean weight of the cars where am = 1.

The null hypothesis H₀ and alternative hypothesis H₁ is given by

Ho :μι – μ2 = 0

against

Hul - 2 + 0

(b) We compute two sample t test.

Two Sample t-test

data: data1 and data2
t = 5.2576, df = 30, p-value = 1.125e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.8304317 1.8853577
sample estimates:
mean of x mean of y
3.768895 2.411000

(i) p-value = 1.125e-05

(ii)A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. It indicates strong evidence against the null hypothesis, as there is less than a 5% probability the null is correct (and the results are random). Therefore, we reject the null hypothesis, and accept the alternative hypothesis

c. .95 percent confidence interval:
0.8304317 1.8853577

(d)

The null hypothesis H₀ and alternative hypothesis H₁ is given by

Ho :μι – μ2 = 0

against

H, :μι – μ2 > 0

Two Sample t-test

data: data1 and data2
t = 5.2576, df = 30, p-value = 5.627e-06
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
0.9195388 Inf
sample estimates:
mean of x mean of y
3.768895 2.411000

(i) p-value = 5.627e-06

(ii) A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. It indicates strong evidence against the null hypothesis, as there is less than a 5% probability the null is correct (and the results are random). Therefore, we reject the null hypothesis, and accept the alternative hypothesis.

The R code is attached.

- 0 X RRGI (64-bit File Edit Packages Windows Help Untitled - R Editor attach tears) detal<- subset (mt cars, "0", "w")) data2c-subset (toare, 1 )) c. test (datal, data2, alcato.sided".var.equal-TRUE) c. test (detal, data2, als greater", var.equal-TRUE)