Question

A national survey indicated that the mean weight of 14-year-old boys was 50 kilograms. In addition, the population was normally distributed. Based on preliminary data, a researcher believes that the average for 14-year-old boys in Maine is different from the national average. To investigate this belief, she randomly samples six 14-year-old boys and records their weights: 44, 47, 45, 48, 49, and 49.

(a) Which statistical test does this scenario call for?

(b) Why that statistical test? Hint: What do we know and what do we not know?

(c) State H0.

(d) State H1.

(e) Please justify your choice of H1.

(f) What are your degrees of freedom (if there are any)?

(g) What is/are the critical value(s) of your test statistic based on $\alpha$ = 0.05

(h) Which of the areas in the figure below (A, B, C, and/or D) reflect the critical region(s)? Select all that apply.

(i) Please calculate the value of each of the following. Hint: Remember, we are now doing inferential statistics.

Sample mean :

Population mean :

Standard deviation :

Standard error:

(i) What is the formula for your test statistic? Hint: It’s probably safer/easier to type it out using parentheses and “ +, -, *, or / ”, but you can write it out in English as well.

(j) What is your obtained value for your test statistic?

(k) What is your p-value?

(l) What is the statistical decision regarding H0?

(m) What is the substantive conclusion? Hint: Be certain that it does not sound like a statistical conclusion.

(n) What is the 95% confidence interval for your estimate of the mean weight of 14-year-old boys in Maine?

(o) In what specific way do the results of (l) and (n) agree?

Answer 1

The problem is solved using basic theory of testing of hypothesis.

Find the solution attached.

Let x = weight of 14-year old boy... Given, XN NCM, 02, MEIR, oro A. researchen believes that the. average weight of 14-year-old boys in Maine different from national average which is 50kg of 6 The research er obtained weights pandomly selected boys (a) Here, have. to perform students t-test ..we. 1 14 Wednesday (b) Here, for We. need to test. samble me an. Form.celly... Here, we Ho. u=50 ag Hi! u £50 dont know population variance need to perform student's o so, we t-test. Ho u claim of national susuey (d) Hi u €50 As the researcher belives the weight is different from national arg.

ce). but Research en believes. that the weight is different from national arg. she doesnt. elaim whether the weight 18 greater than less than the national arg. so, we choose an a hence, both tailed test He u €50 XN N(M, 02 =) X N N ( M, 0²/n) .. un (x-u) N CO, 1) (n-1)52 02 where, Sample in (x-4) lo ~tnal xanal 18 Sunday S sd. (-1) 52 | m-) sm (x-4) tn) S degrees. of freedom. Hence, von Here (n-1) 6 m. sample size= 6-12 5. df =

19 21 18 25 20 27 28 29 30 31 26 ci) sample. mean : x = IŠ Xi Ho Ž xi 6 47 = u 50 national population mean. as. stated by std deviation. survey a) 2 s crei- 47) I (xi- a -los V 2 . 09. Sud 26 Std error van Monday 86 The. formula for test - Statistie t. rnlx-M) S size sample n = y = Sample mean. S. u z mean √6 ( 47 - 50) (j) Value of test stat 2009 - 3.51 Sd. of ulation t.

29 (K) þ-value 2emes) 2 Pct > Itcobs) 1] obtained using a. This is t-table standard a þ-value 0.01722 I can be obtained from R- programming also using 'to test' comm and... but its suggested to lools at amy stand and t-table (d) Here þ-value a 0.01722 critical value X 0.05. Friday p-value < we om Hence, reject Ho at 57. level of significance. The mean weight of 14-year-old of Maine differs national significantly from the average: 95. CI is : (F-2 to/2; x+ 2 tale 44.79 49:20 boys, (n)

(6) om The 95). CI does not contain Ho value 50. sample mean differs fr 50. so we reject Ho. result (U) m) agrees Ho is rejected at 51. level of significance and that