Question

Q8. In an amusement park, a booth costs 2$ to play and rewards participants 10$ if they can shoot within 1cm of some target with a bow and arrow. Before doing his advertisement, the manager wants to make sure that people, on average, shoot at least 2cm away from the target. She randomly selects 9 participants to try the game. Their mean distance from the target was 3cm, with a sample standard deviation of 3cm. a) What would be a sensible null hypothesis (Ho) for the manager to test? Ho: j = 2cm b) Against what alternative hypothesis (Ha) should the manager test the null hypothesis? Haili > 2cm c) What is the p-value for this test? 0.2>p-val>.15 d) Should you reject the null at a level of significance of 0.05? Why? No, because the p-value is greater than 0.05. e) If she wants to build a 95% confidence interval with a margin of error of lcm, at least how many people should she have in her sample? 38 (obtained using t(30), using t(40) also ok)

The answers are provided in red but I would like to see the calculations and steps about how to get the answers, Thanks

Answer 1

for the given data since the manager wants to check that the shoot is at least 2cm away from the target means he wants to check the shoot if it is greater than 2cm, the Hypotheses for this test will be:

a) The hypotheses are:

Ho: μ = 2 Ha: μ> 2

so, the Null hypothesis against which he will test will be:

Ho:μ = 2

b) The Alternate hypothesis will be:

Ho : > 2

based on the hypothesis it will be a one-tailed test. Now since the sample size is less than 30 and the population standard deviation is unknown hence we will do a t-test instead of a Z test.

The test statistic is calculated as:

t= X - μo 8/νη 3 - 2 3/ 9

given that the sample size is 9 hence the degree of freedom will be n-1 which is in this case 9-1=8

c) P-value:

The P-value is calculated at the t-statistic and also a degree of freedom that we have calculated above by using either by the T table or by the excel formula which is =1-T.DIST(1,9,TRUE).

The P-value yielded is 0.1717

so, 0.20>P-val>0.15

d) Now at 0.05 level of significance to reject the Null hypothesis we compare the value of P-value with the level of significance which is 0.05 in this case. if the P-value <0.05 then we can say that we can reject the Null but if P-value>0.05 we cannot reject the null hypothesis.

e) Now since we do not know the sample size and margin of error E is given here as1 cm, so we can assume that the sample size could be of any size so we will use the Z statistic to calculate the minimum sample size.

N = 2 *

here the Zc is calculated again by excel formula =NORM.S.INV(0.975) which gives us Z=1.96, now the question is why we are using 0.975 since we have confidence level is 0.95. The answer is because the confidence level 0.95 is in the between of the normal bell-shaped curve but the critical value or probability using excel or by Chart always gives from 0 that is the left side of the bell-shaped curve.

N = 1.962 * = 34.572 35