Question

Problem 1. (1 point) Weather tracking records indicate that in the previous year, the mean daily-maximum temperature in Calgary over the months of May, June, July, and August is 26.6 Celsius. Remaining time: 86:41 (min:sec) The data appearing in column C1 of the Download.csv file resulted from a random sample of 18 days taken between May 1 and August 31. For each day, the daily maximum temperature in degrees Celsius was observed. The data als 26.6, 22.4, 27.3, 19.5, 11, 18.3, 18.8, 13.5, 25.4, 19.1, 26.5, 16.8, 15.5, 13, 24.2, 19.8, 29.5, 7.2 Report the p-value of the normality of the data. P-value= 0.8045 (use three decimals in your answer) (a) You wish to test the hypothesis that it was statistically warmer this past May, June, July, and August than the previous year. Select the appropriate statistical hypotheses. A. H. : H = 26.6 HA:41 # 26.6 B. H. : 4 < 26.6 HA:> 26.6 OC.H: X < 26.6 HA:X > 26.6 D. H, : X > 26.6 HA: X < 26.6 E. Ho : X = 26.6 HA :X # 26.6 OF. H, :> 26.6 HA:H <26.6 OG. H. : 4 > 26.6 HA: 4 < 26.6 (b) The mean and standard deviation of these data were found to be X = 19.689 and S = 6.194. Compute the value of the test statistic. Test Statistic = (use three decimals in your answer) (c) Using the technology available to you compute the P-value. P-value = (use four decimals in your answer)

how to do this question using r studios

Answer 1

The output from R studio is ,

> X=c(26.6,22.4,27.3,19.5,11,18.3,18.8,13.5,25.4,19.1,26.5,16.8,15.5,13,24.2,19.8,29.5,7.2)
> X
[1] 26.6 22.4 27.3 19.5 11.0 18.3 18.8 13.5 25.4 19.1 26.5 16.8 15.5 13.0 24.2
[16] 19.8 29.5 7.2
> t.test(X,mu=26.6)

One Sample t-test

data: X
t = -4.734, df = 17, p-value = 0.0001919
alternative hypothesis: true mean is not equal to 26.6
95 percent confidence interval:
16.60877 22.76900
sample estimates:
mean of x
19.68889
Hypothesis :
H, :μ = 26.6.
VS
H, :μ = 26.6

Test statistic = -4.734

p-value=0.0001919$\approx$0.0002