Question

The accompanying table contains data on the​ weight, in​ grams, of a sample of 50 tea bags produced during an​ eight-hour shift. Complete parts​ (a) through​ (d).

a. Is there evidence that the mean amount of tea per bag is different from 5.5 ​grams? (Use alpha = 0.05​.)
ALSO STATE THE NULL AND ALTERNATIVE HYPOTHESIS:

DETERMINE TEST STATISTIC:
The test statistic is
​(Round to two decimal places as​ needed.)

FIND THE P-VALUE:
​p-value =
​(Round to three decimal places as​ needed.)

STATE THE CONCLUSION:
Upper H 0. There is ▼ sufficient OR insufficient evidence to conclude that the mean difference is not equal to 5.5 inches.

b. Construct a 95​% confidence interval estimate of the population mean amount of tea per bag. Interpret this interval.
The 95​% confidence interval is:

​(Round to four decimal places as​ needed.)

Interpret the 95​% confidence interval. Choose the correct answer below.
A. Do not reject Upper H 0 because the hypothesized mean is contained within the confidence interval.
B. Reject Upper H 0 because the hypothesized mean is not contained within the confidence interval.
C. Reject Upper H 0 because the hypothesized mean is contained within the confidence interval.
D. Do not reject Upper H 0 because the hypothesized mean is not contained within the confidence interval.

c. Compare the conclusions reached in​ (a) and​ (b). Choose the correct answer below.
A. The confidence interval and hypothesis test both show that there is sufficient evidence that the mean amount of tea per bag is different from 5.5 grams.
B. The confidence interval shows insufficient evidence while the hypothesis test shows sufficient evidence that the mean amount of tea per bag is different from 5.5 grams.
C. The confidence interval shows sufficient evidence while the hypothesis test shows insufficient evidence that the mean amount of tea per bag is different from 5.5 grams.
D. The confidence interval and hypothesis test both show that there is insufficient evidence that the mean amount of tea per bag is different from 5.5 grams.

5.65
5.44
5.41
5.41
5.54
5.34
5.54
5.45
5.51
5.42
5.55
5.41
5.51
5.53
5.54
5.64
5.54
5.47
5.44
5.51
5.46
5.42
5.46
5.62
5.53
5.31
5.68
5.28
5.48
5.56
5.77
5.56
5.41
5.58
5.59
5.51
5.32
5.48
5.53
5.57
5.61
5.44
5.43
5.23
5.55
5.63
5.48
5.57
5.67
5.35

Answer 1

$\mu$ : mean amount of tea per bag

Null hypothesis : H_o : $\mu$ =5.5

Alternate Hypothesis : H_a : $\mu$ $\neq$5.5

Two tailed test.

Hypothesized mean : $\mu _{o}$ = 5.5

x: sample data of weight in grams of tea in a bag

Sample mean : sample mean weight of tea per bag

$\overline{x}=\frac{\Sigma x}{n}$

n : Sample size = 50

sample standard deviation :s

(3 - 72 = v n - 1

	x	(x-$\overline{x}$)	(x-$\overline{x}$)2
	5.65	0.1514	0.022922
	5.44	-0.059	0.003434
	5.41	-0.089	0.007850
	5.41	-0.089	0.007850
	5.54	0.0414	0.001714
	5.34	-0.159	0.025154
	5.54	0.0414	0.001714
	5.45	-0.049	0.002362
	5.51	0.0114	0.000130
	5.42	-0.079	0.006178
	5.55	0.0514	0.002642
	5.41	-0.089	0.007850
	5.51	0.0114	0.000130
	5.53	0.0314	0.000986
	5.54	0.0414	0.001714
	5.64	0.1414	0.019994
	5.54	0.0414	0.001714
	5.47	-0.029	0.000818
	5.44	-0.059	0.003434
	5.51	0.0114	0.000130
	5.46	-0.039	0.001490
	5.42	-0.079	0.006178
	5.46	-0.039	0.001490
	5.62	0.1214	0.014738
	5.53	0.0314	0.000986
	5.31	-0.189	0.035570
	5.68	0.1814	0.032906
	5.28	-0.219	0.047786
	5.48	-0.019	0.000346
	5.56	0.0614	0.003770
	5.77	0.2714	0.073658
	5.56	0.0614	0.003770
	5.41	-0.089	0.007850
	5.58	0.0814	0.006626
	5.59	0.0914	0.008354
	5.51	0.0114	0.000130
	5.32	-0.179	0.031898
	5.48	-0.019	0.000346
	5.53	0.0314	0.000986
	5.57	0.0714	0.005098
	5.61	0.1114	0.012410
	5.44	-0.059	0.003434
	5.43	-0.069	0.004706
	5.23	-0.269	0.072146
	5.55	0.0514	0.002642
	5.63	0.1314	0.017266
	5.48	-0.019	0.000346
	5.57	0.0714	0.005098
	5.67	0.1714	0.029378
	5.35	-0.149	0.022082
Total	274.93		Σ(x – ΤΣ =0.572202
Mean: $\overline{x}$ =274.93/50	5.4986

(3 - 72 = v n - 1

(x - 72 n-1 0.572202 -= 0.1081 V 50 -1 V

Test Statistic : tstat = T-HO s/n = (5.4986 – 5.5) 0.1081/50 -0.0014 0.0153

The test statistic = -0.0915

Fro two tailed test:

P - Value = P(t < -tstat) + P(t > tstat) = 2 x P(t > 0.0915)

for 49 degrees of freedom , P(t>0.0915) = 0.4637

p-value = 2xP(t>0.0915) = 2x0.4637=0.9274

p-value = 0.9274

As P-Value i.e. is greater than Level of significance i.e (P-value:0.9275 > 0.05:Level of significance); Fail to Reject Null Hypothesis
There is insufficient evidence to conclude that the mean difference is not equal to 5.5 inches

b. Construct a 95​% confidence interval estimate of the population mean amount of tea per bag. Interpret this interval.

confidence interval estimate of the population mean

s : Tta/2.n-1 XP vn

$\alpha$ for 95% confidence level =(100-95)/100=0.05

$\alpha$/2 =0.05/2=0.025

+0.025,49 = 2.0096

95​% confidence interval estimate of the population mean amount of tea per bag

I to.025,49 X == 5.4986 = 2.0096 x 0.1081 -= 5.4986 – 2.0096 x 0.0153 V50 Vn2 = 5.4986 + 0.0307 = (5.4679,5.5293)

The 95​% confidence interval is:(5.4679,5.5293)

Interpret the 95​% confidence interval. Choose the correct answer below.
A. Do not reject H₀ because the hypothesized mean is contained within the confidence interval.

c. Compare the conclusions reached in​ (a) and​ (b). Choose the correct answer below.

D. The confidence interval and hypothesis test both show that there is insufficient evidence that the mean amount of tea per bag is different from 5.5 grams.