Question

Temperatures (in Celsius) of a city were observed for 10 random days to test Ho: u = 15 versus Hau # 15. A student changed the data to Fahrenheit. The conversion formula for changing Celsius (C) to Fahrenheit (F) is: F = (975)C + 32. If using the data converted into Fahrenheit to test Ho: u = 59 versus Ha: # 59, which of the following statements is TRUE about the test statistic, degrees of freedom (df), and conclusion at the 5% significance level? Select one: O a. The test statistic is the same, but the df and conclusion are different. cross out b. The test statistic, df, and conclusion are all different. cross out O c. The test statistic is different, but the df and conclusion are the same. cross out cross out d. The test statistic and df are different, but the conclusion is the same. e. The test statistic, df, and conclusion are all exactly the same. cross out

Answer 1

Temperatures (incelsius) of a city were observed for 10 random days to test, Hoi M=15 vs Haiv46159 A Student Change the data to Fahrenheit The conversion formula for changing ceisius (e) to Fahrenheit (Flis: F = (a +32. If using the denta converted into Fahrenheit to test Hoi (4=59 vs HAY M759. Now, for this test our appropriate test Statistic is given by, n (X-MO NA(n-u . Tewhere x sample mean, se samples g= number of random sample Mo Hypothesized mean]

NOW Cve convert the hypothesized mean (celsius) to hypothesized mean Fahrenheit, by the formula, * F319 32+32 YF: 2 x t +32 sy Fr 27+32= 59.
And here given that after converting the data our hypothesized mean is 59. That is our null and alternative hypothesis is exactly same.

Define, $\bar{c}=$ Sample mean for Celsius , $s_c=$ Sample sd for Celsius.

Test statistic in Celsius:-

$=>T_1=\frac{\sqrt{10}(\bar{c}-15)}{s_c}$

Now we convert $\bar{c}$ and $s_c$ in Fahrenheit and we get,

Fahrenheit and,

$=>\bar{c}=(9/5)\bar{c}+32$ Fahrenheit

Test statistic in Fahrenheit:-

$=>T_2=\frac{\sqrt{10}(\frac{9}{5}\bar{c}+32-59)}{\frac{9}{5}\times s_c}$

$=>T_2=\frac{\sqrt{10}(\frac{9}{5}\bar{c}-27)}{\frac{9}{5}\times s_c}=\frac{\sqrt{10}\times\frac{9}{5}(\bar{c}-15)}{\frac{9}{5}\times s_c}$

$=>T_2=\frac{\sqrt{10}(\bar{c}-15)}{ s_c}=T_1$

=> So we get test statistic is exactly same in Celsius and Fahrenheit.

=> And both the cases (Celsius and Fahrenheit) have same number of random sample 10. So df = (10-1) = 9 is also same in the both the cases (Celsius and Fahrenheit).

=> And we know that if test statistic and df is same then conclusion is also same. Because we draw conclusions based on test statistic and df.

Answer:- Correct answer is " Option e. " (The test statistics, df and conclusion are all exactly the same)