Question

8of 14 (11 complete) HW Sco 9.2.34-T on the weight,in grams, of a sample of 50 tea bags produced during an eight-hour shift Complete parts (a) hrough (d). EEE Click the icon to view the data table. State the null and alternative hypotheses Ho : 5.5 H H 55 Type integers or decimals.) Determine the test statistic The test statistic is 0.21 Round to two decimal places as needed.) Find the p-value. More Into Tea Bag Weight (in grams) 64 545 5.43 5.38 554 534 556 547 553 539 5.55 5.38 552 554 557 561 558 547 5.45 55 5.49 5.39 548 562 554 53 569 529 549 555 5.76 556 5.43 558 5.56 551 531 540 554 558 5.62 5.43 5.45 5.25 5.58 563 551 557 567 536 p-vakue- 0 834 (Round to three decimal places as needec State the conclusion. Print Done Do not reject Ho There is insufficient evidence to concle b. Construct a 96% condence interval essmase of the population mean amount onea per bag Interpret this nerval. The 95% confidence interval is--$15 Enter your answer in the edit fields and then click Check Answer Clear All remaining esc lab KL

Answer 1

1) We will set up the null hypothesis

Ho : μ = 5.5

$H_{1}:\mu \neq 5.5$

2) Under the null hypothesis the test statistics is

Vn

where
5.64 + 5.55 + 50 + 5.36 「Z'İ = 5.5032 に!

$S=\sqrt{\frac{\sum_{i=1}^{n}( x_{i}-\bar{x})^{2}}{n-1}}$

$S=\sqrt{\frac{(5.64-5.5032)^2+(5.55-5.5032)^2+...+(5.36-5.5032)^2}{50-1}}$

S- 0.1075

$t=\frac{5.5032-5.5}{\frac{0.1075}{\sqrt{50}}}$

$t=0.21$

3) The corresponding  

4) Since calculated which is greater then 0.05 hence we will accept our null hypothesis and concludes that
μ = 5.5
.

5) The formula used for the construction of confidence interval is given below

$C.I=\bar{x}\pm \frac{t_{\frac{\alpha }{2}}S}{\sqrt{n}}$ Where $t_{\frac{\alpha }{2}}$ is the critical value at 0.05 significance level with 49 df = 2.011.

$C.I=5.5032\pm \frac{2.011*0.1075}{\sqrt{50}}$