Question

Critical Values of the t-Distribution (Page 1) Arandom sample of b3 bags of white cheddar popcoin weighed, on average, 532 ounces with a standard deviation of 0.25 ounce Test the hypothesis that i=55 Dunces against the aternative hypothesis, 1455 ounces, at the 0.10 level of significance Click here to via CA 1 of the table of critical values of the tristribution Click here to view age 2 of the table of critical values of the t-distribution Identify the critical region. Select the correct choice below and fill in the ansatr box(es) to complete your choice. Round to two decimal places as needed.) (L. 0.10 8. 0.325 25 1275 1 2 31 12 0277 0221 1.63 1. or 1> 0.20 1.376 11 0.98 0911 0.920 0.906 0896 0469 0.853 0.15 1953 1.346 1.190 114 11:14 1119 1106 1100 1 100 OA 1 B. 1> OC 1 2571 2.47 0.53 1. 1.416 0.05 0.114 1991 25 2.13 2. 1.948 1.96 1.800 13 12 1.296 1.7 1.77 1.761 a OSB 2 2.22 BE LR 1.888 LITZ 5.00 1361 Find the test statistio. 0.59 6 6 25 7 5 9 021 10 11 12 259 13 14958 15 028 10 0258 0253 1 1s 0.257 19 0.257 20 21 257 0896 1 083 1043 0.870 LUS 069 1076 1 ITA Round to two decimal places as needed) 1.336 1.130 1.3 1141 1.99 2.201 2179 2100 2135 21:11 120 2110 2101 What is the appropriate conclusion for this test? OLI OA Reject H and conclude that the average weight of the bags of white cheddar popcom is not significantly less than 5.5 ounces. OB. Reject Ho and conclude that the average weight of the bags of white cheddar popcom is significantly less than 5.5 ounces OC. Do not reject H and conclude that the average weight of the bags of white cheddar popcom is not significantly less than 5.5 cunces. OD. Do not reject He and conclude that the average weight of the bags of white cheddar popcom is significantly less than 5.5 ounces 086 108 1056 1054 1.820 0.50 1053 1.29 1 1.121 0.85€ 1.050 1.519 0857 1 EVO 1.1 10 0856 1056 1.316 LAST 05 nu 1.313 0554 1655 1311 2016 1.746 1.740 1.734 1.729 1.725 1.721 1.711 1.734 1.711 1.708 1706 1.705 1.no 1090 LOT 2 2! 28 2. 24 25 26 0256 0.256 பொம்ப 0331 0531 0.631 3 30 25 22 2008 2035 203 LAS Print Done i Cntical values of the Distribution (Page 2) х 4317 4029 22 1165 20 5. 5. 5041 INI 2160 10.000 4.ST 2.ST 33 2 3.100 3106 8.055 3012 2.977 2.901 2.771 2.921 8.438 3336 390 3959 21 2174 2.158 9.00 0 3.143 7 2.517 2.715 2.959 2419 9 21 2.764 11 2.330 2.091 2.718 12 2.30G 2.661 132.242 2.66 2.450 2.254 2.024 15 2.319 2.397 16 25 17 2.56 18 2.216 19 2.205 2.836 2.450 20 2.197 2.5 21 2.318 2.153 2 4 2313 2 24 2317 2.31 2150 26 2.12 2.996 2.119 2.231 2.IT 2286 20 2.150 2.462 30 2.147 an 20 2293 2.350 120 2.078 2.196 2.358 2005 2.170 0.0 0.015 COL 27 2 2 2.674 2.661 260 2 1.318 121 1140 1903 1012 1.96 1 R3 1.500 1819 3.790 2018 2 23 2831 2818 2017 3.119 13 S 200 1.77 2.500 3078 306 205 3041 3058 3.030 2991 28 2.76 2.771 2.1 2.756 2.750 2.14 2010 2017 2.676 2.581 2 3.614 3.560 3.50 22551 2.850 2.469 2.452 C.ORTE 0.0025 0.0005 Print Done

Answer 1

The value provided in the above question are as below

sample size = = 63

sample mean = $\bar{x}$ = 5.32

standard deviation = = 0.25

S

population mean = $\mu$ = 5.5

$\alpha$ = 0.10

The null and alternative hypothesis is

$H_{0} :\mu =5.5$

$H_{1} :\mu <5.5$

The alternative hypothesis indicate that the given test left tailed.

a) Identify the critical region. Select the correct choice

Answer :- t < -1.30

b) Find the test statistic

Answer :- -5.71

What is appropriate conclusion for this test

Answer :- B. Reject Ho and conclude that the average weight of the bags of white cheddar popcorn is significantly less than 5.5 ounces.

The complete Solution of above part is as below

a) We find critical region using t table

The required value is degrees of freedom = n - 1 = 63 - 1 = 62 and $\alpha$ = 0.10

We find the value in table which degrees of freedom = 62 (choose approximately = 60 in t table) and $\alpha$ = 0.10

The value is 1.296 using 2 decimal places = 1.30 but hypothesis is left tailed then we take negative sign to critical region

Answer :-t < -1.30

b) The test statistic is calculate using formula as below

$t=\frac{\bar{x}-\mu }{s/\sqrt{n}}$

  $=\frac{5.32-5.5}{0.25/\sqrt{63}}$

$t=-5.71482 \approx -5.71$ (Round up to 2 decimal places)

$t=-5.71$

test statistic = -5.71

What is appropriate conclusion for this test

Answer :- We comparing test statistic with critical value and take decision of reject or do not reject Ho

Here, test statistic = -5.71 < critical value = -1.30

The conclusion is

B. Reject Ho and conclude that the average weight of the bags of white cheddar popcorn is significantly less than 5.5 ounces.