Question

Match the following words to their correct definitions. Write the correct letter in the space next to the definition. (1 point each).

Note: You will not use all the words.

______a subset of the population we are interested in studying

______The difference between the sample measure and the corresponding population measure.

______the total set of subjects that are being studied

______The number of standard deviations away from the mean a particular data point is _____

______A normal distribution with a mean of zero and a standard deviation of one.

______a statement that there is a difference between a parameter and some claimed value.

______the maximum probability of committing a type I error.

A. Sample        

B. Population        

C. Parameter        

D. Statistic        

E. Z-score   

G. Sampling Error      

I. Standard Normal Distribution       

K. Type I Error        

L. Type II Error        

N. Significance Level        

O. Null Hypothesis        

P.  Alternative Hypothesis

22.  (3 points) H₀: µ ≤ 15 versus H₁: µ > 15, α = 0.05, x = 17, s = 8, n = 29

test statistic _________            p-value _________      Decision (circle one)       Reject the H₀   or   Fail to reject the H₀

23. (3 points) H₀: µ ≥ 205 versus H₁: µ < 205, α = 0.05, x = 198, σ = 20, n = 25

test statistic _________            p-value _________      Decision (circle one)       Reject the H₀    or  Fail to reject the H₀

24.   (3 points) The average weight of a package of rolled oats is supposed to be at least 18 ounces. A sample of 18 packages shows a mean of 17.78 ounces with a standard deviation of 0.41 ounces. The calculated test statistic is t = -2.28 (left tail test) and the corresponding p-value = 0.018.

Based on the results above, is the sample mean (17.78 ounces) smaller than the specification (18 ounces) at the 0.05 level of significance? (Check one of the following responses):

A. Yes, the sample mean is significantly smaller than the specification at 0.05 level of significance.

B. No, the sample mean is not significantly smaller than the specification at 0.05 level of significance.

C. Yes, the p-value is greater than the alpha level.

D. No, the p-value is lower than 0.01

ACME Manufacturing claims that its cell phone batteries last more than 32 hours on average in a certain type of cell phone. Tests on a random sample of 18 batteries showed a mean battery life of 37.8 hours with a population standard deviation of 10 hours. Is the mean battery life greater than the 32 hour claim? Answer the following questions using a significance level of alpha = 0.05.

25. (2 points) Write the null and alternative hypotheses:

H₀:

H₁:

26. (2 points) In the box above, show the formula for calculating the appropriate test statistic. Include actual numbers in the formula, for example, z or t = (32-32)/32.

27. (2 points) Calculate the appropriate test statistic:  _____________

29. (3 point) Using a right tailed test, is the average battery life significantly greater than the claim of 32 hours? Explain.

Answer 1

  Sample is a subset of the population we are interested in studying.

  Sampling error is the difference between the sample measure and the corresponding population measure.

Population is the total set of objects that are being studied.

Z score is the number of standard deviations away from the mean a particular data point is.

Standard Normal distribution is a Normal distribution with a mean of zero and a standard deviation of one.

Alternative hypothesis is the statement that there is a difference between the parameter and the claimed value.

Significance level is the maximum probability of committing a Type I error.

22> To test H₀: $\dpi{100} \mu \leq 15$ ag H₁: $\dpi{100} \mu > 15$

Level of significance , $\dpi{100} \alpha$ = 0.05

Sample size, n =29

Sample standard deviation s =8

X = 17

Test statistic T = $\dpi{100} \sqrt{29}$ (17 -15)/8 = 1.346 (correct up to 3 decimal places)

Note that, T $\dpi{100} \sim$ t_n-1=28  

p-value= 0.09455 (obtained using the t table given in Biometrica)

Since the p value 0.09455 is greater than the level of significance 0.05, we fail to reject H₀ as far as the given data is concerned

23>  

To test H₀:$\dpi{100} \mu \geq 205$  ag H₁: $\dpi{100} \mu < 205$

Level of significance , $\dpi{100} \alpha$ = 0.05

Sample size, n =25

Population standard deviation $\dpi{100} \sigma$ = 20

X = 198

   Test statistic Z = $\dpi{100} \sqrt{25}$ (198 -205)/20= -1.75

Note that, Z $\dpi{100} \sim$ N(0,1)

p value = 0.04 ( obtained using the Z table given in Boiometrica)

Since the p value 0.04 is less than the level of significance 0.05, we reject H₀ as far as the given data is concerned.

24> Since the p value obtained in the test 0.018 is less than the level of significance 0.05, we reject the null hypothesis  and  conclude that the sample mean is significantly smaller than the specification at the 5% level of significance.

25>    Null Hypothesis : Mean battery life is 32 hours.

Alternative hypothesis : Mean battery life is greater than 32 hours.

26>    To test H₀ : $\dpi{100} \mu =32$ ag H₁: $\dpi{100} \mu > 32$

Sample size n = 18

Population standard deviation  $\dpi{100} \sigma$ = 10

  Sample mean $\dpi{100} \bar{X}$ = 37.8

level of significance $\dpi{100} \alpha$ = 0.05

   Test statistic Z = $\dpi{100} \sqrt{n}$ ($\dpi{100} \bar{X}$ - 32) / $\dpi{100} \sigma$   $\dpi{100} \sim$ N(0,1)

27>    Test statistic Z =     $\dpi{100} \sqrt{18 }$ ( 37.8 - 32)/ 10 = 2.4607 ( correct up to 4 decimal places)

28> Note that , Z $\dpi{100} \sim$ N(0,1)

Z_{$\dpi{100} \alpha$}= Z_0.05 = 0.5199388

Clearly, Z > Z_{$\dpi{100} \alpha$.}

So we reject the null hypothesis and conclude that the mean battery life is greater than 32 as par as the given data is concerned.