Please give answer to number h. We would like to conduct a test of significance
ID: 3128949 • Letter: P
Question
Please give answer to number h.
We would like to conduct a test of significance at the 10% level of significance to determine whether the mean daily caloric intake of adult males is differs from that of females. We take random samples of ten men and eight women and record the number of calories they consume on a particular day. The data are shown in the table below:
Men 2575 2210 2995 2950 2745 3100 2490 2530 2310 3510
Women 2440 2090 1935 2390 2020 2325 2125 2350
We calculate xM=2741.5, sM=398.8, xW=2209.4, sW=189.5 .
(a) What are the degrees of freedom we will use when looking up critical values and P-values on Table 3?
(b) Calculate a 90% confidence interval for MW, the difference in mean caloric intake between males and females.
(c) What is the value of the test statistic for the appropriate test of significance?
(d) What is the P-value for the appropriate test of significance?
(e) Verify your results by conducting the appropriate hypothesis test in JMP. To do this, create a column called Calories and enter all 18 data values in this column (all 10 data values for men and then all 8 data values for women). Create a second column called Gender and type M in the first 10 rows (beside the data values for the men) and W in the next 8 rows (beside the data values for the women). Go to Analyze > Fit Y by X. Choose Calories as Y and Gender as X. Click OK. Now, if you want to conduct a pooled two-sample t test, under the red arrow, select Means, Anova, Pooled t. If you want to conduct a conservative (unpooled) two sample t test, under the red arrow, select t Test.
What is the exact P-value of the test?
(f) Suppose we had instead used the critical value method to conduct the test. What would be the decision rule?
(g) What is the correct conclusion for the test?
(h) Could the confidence interval in (b) be used to conduct the hypothesis test? Why or why not? If the interval could be used to conduct the test, what would the conclusion be, and why?
Explanation / Answer
The data is given for daily caloric intake of adult males and women. for that study we collect the sample of size 10 and 8 respectively . we use the R- software for calculation
we calculate the mean and sd of the given sample is as follows ;
men<-c(2575,2210,2995,2950,2745,3100,2490,2530,2310,3510)
women<-c(2440,2090,1935,2390,2020,2325,2125,2350)
> mean(men)
[1] 2741.5
> mean(women)
[1] 2209.375
> sd(men)
[1] 398.7832
> sd(women)
[1] 189.5377
a. at first we test H0: The mean daily caloric intake of adult males is differs from that of females. for that statements we use the two sample t-test for unequal size sample. before doing this test at first see variance is equal or not if not than df of the t- statistics is different . the testing results is as follows with code
t.test(men,women,alternative = c("two.sided"),paired = FALSE, var.equal = FALSE, conf.level = 0.90)
Welch Two Sample t-test
data: men and women
t = 3.7262, df = 13.424, p-value = 0.002412
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
279.8359 784.4141
sample estimates:
mean of x mean of y
2741.500 2209.375
hence the df is 13.242 in the test statistics. is used for calculating the p- value.
b. the confidence interval of the MW, the difference in mean caloric intake between males and females at 90 % confidence interval is 22029.375 to 2741.5.
c. the calculated t- statistics value is t = 3.7262.
d. the p- value is = 0.002412 , here p- value is less than we say that the mean daily caloric intake of adult males is differs from that of females
Note. we have no JMP software so i left the remaing subpart of this question