Question

1. A random sample of 30 boxes of bolts was purchased at a local hardware store and the average number of bolts per box was 52. It is known that the population standard deviation of bolts per box is 6 bolts. a Is this a z or t-distribution? Why? b. Find and interpret a 95% confidence interval to estimate the average number of bolts per box for all boxes in the population. Round to 3 decimal places. C What would happen to this interval if the number of boxes in the random sample was increased?

Answer 1

Solution:

Question 1)

Given:

Sample size = n = 30

Sample mean = = 52

T

Population standard deviation = $\sigma$ = 6

Part a)

Since population standard deviation is known, we use z test.

Part b)

Confidence level = c = 95%

Formula:

T-E << 1 + E)

where

E = ze X o/Vn

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

2 .00 .01 .02 03 .04 .05 .06 .07 .08 09 0.0 0.1 0.2 0.3 0.4 0.5 .5139 .536 .6026 .6406 .672 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 .5000 .5398 .5793 .6179 .6554 .6915 .7257 .7580 .7881 .8159 .8413 .8643 .8849 .9032 .9192 .9332 .9452 .9554 .9641 0743 .5040 .5438 .5832 .6217 .6591 .6950 7291 7611 .7910 .8186 .8438 .8665 .8869 .9049 .9207 .9345 .9463 .9564 .9649 .9719 .5080 .5478 .5871 .6255 .6628 .6985 .7324 7642 7939 .8212 .8461 .8686 .8888 .9066 .9222 .9357 .9474 .9573 .9656 .9726 .5120 .5517 .5910 .6293 .6664 7019 .7357 7673 .7967 .8238 .8485 .8708 .8907 .9082 .9236 .9370 .9484 .9582 .9664 .5160 .5557 .5948 .6331 .6700 .7054 .7389 .7704 .7995 .8264 .8508 .8729 .8925 .9099 .9251 .9382 .9495 .9591 .9671 .5199 .5596 .5987 .6368 .6736 .7088 .7422 .7734 .8023 .8289 .8531 .8749 .8944 .9115 .9265 .9394 .9505 .9599 .9678 0744 .123 1454 764 B051 B315 .6554 .8770 .8962 .9131 .9279 .9406 .9515 .9508 .9686 .9750 .5279 .5675 .6064 .6443 .6808 .7157 7485 7794 .8078 .8340 .8577 .8790 .8980 .9147 .9292 .9418 .9525 .9616 .9693 .9756 .5319 .5714 .6103 .5480 .6844 7190 .7517 .7823 .8106 .8365 .8599 .8810 .8997 .9162 .9306 .9429 .9535 .9625 .9699 .9761 .5359 .5753 .6141 .6517 .6879 7224 .7549 .7852 .8133 .8389 .8621 .8830 .9015 .9177 .9319 .9441 .9545 .9633 .9706 .9767 בכל סבלם.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

thus

E = ze X o/Vn

$E=1.96 \times 6 / \sqrt{30}$

$E=1.96 \times 6 / 5.477226$

thus

T-E << 1 + E)

$( 52-2.147 \: \: < \mu < \: \: 52+2.147 )$

$\mathbf{{\color{DarkOrange} ( 49.853 \: \: < \mu < \: \: 54.147 )}}$

Part c)

Margin of error is given by:

E = ze X o/Vn

Since Margin of error is inversely proportional to sample size n, as n increases, margin of error decreases and hence length of confidence interval is also decreases.

Question 2)

Given:

Sample size = n = 44

Sample mean = = $13,445

T

Sample standard deviation = s= $1500

Population mean = $\mu =\$ 12,837$

Part a)

Since population standard deviation is unknown, we use t test.

Part b)

Hypothesis in symbols:

$H_{0}: \mu =\$ 12,837$ Vs $H_{a}: \mu > \$ 12,837$

Hypothesis in words:

H₀: Ph.D. students in a large university in the state earn an average salary $12,837

Vs

H_a: Ph.D. students in a large university in the state earn an average salary more than $12,837

Part c)

Given:

Mean Sample Mean Std. Err. DF T-Stat P-value д 13445 226.13351 43 2.6886772 0.0051

Decision Rule:
Reject null hypothesis H0, if P-value < 0.01 level of significance, otherwise we fail to reject H0.

Since P-value = 0.0051 < 0.01 level of significance, we reject null hypothesis H0.

Part d)

Conclusion:

At 0.01 level of significance, we have sufficient evidence to conclude that: Ph.D. students in a large university in the state earn an average salary more than $12,837.

Thus the dean is correct.