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3. Supercavitation is a propulsion technology for undersea vehicles that can gre

ID: 3054810 • Letter: 3

Question

3. Supercavitation is a propulsion technology for undersea vehicles that can greatly increase their speed. It occurs above approximately 50 ?????1, when pressure drops sufficiently to allow the water to dissociate into water vapor, forming a gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle, supercavitation is said to occur. Eight tests were conducted on a scale model of an undersea vehicle in a towing basin with the average observed speed ???=102.2 ?????1. Assume that speed is Normally distributed with known standard deviation ??=4 ?????1. You need to test if the average speed is less than 100 ?????1.

a. Write down the null and alternative hypotheses to be tested.

b. Test the hypotheses at ??=0.05, using the test statistic approach. What is your decision on null hypothesis (??0)? Interpret the test results.

c. Construct a 95% confidence interval for the mean speed. Test the hypotheses using the confidence interval method. Do you get the same answer as part (b)?

d. What is the ?? ?????????? for the test? Test the hypotheses using the ?? ??????????. Do you get the same answer as parts (b) and (c)?

Explanation / Answer

a)

Hypothesis:

H0 : mu = 100

Ha : mu < 100

b)

t = ( x - mean) / (s/sqrt(n))

= (102.2 - 100)/ ( 4/sqrt(8))

= 1.5556

Rejection Region for Lower-Tailed Z Test (H1: ? < ?0 ) with ? =0.05

The decision rule is: Reject H0 if Z < 1.645.

we reject the null hypothesis.

c)

CI for = 95%

n = 8

mean = 102.2

t-value of 95% CI = 2.3646

std. dev. = 4

SE = std.dev./sqrt(n)

= 4/sqrt(8)

= 0.35188

ME = t*SE

= 2.3646 * 0.35188

= 0.83207

Lower Limit = Mean - ME

= 102.2 - 0.83207

= 101.36793

Upper Limit = Mean + ME

= 102.2 + 0.83207

= 103.03207

95% CI (101.3679 , 103.0321 )

As hypothesis does not contain null hypothesis value so we can reject the null hypothesis.

d)

P value is calculated using t = 1.5556 at 0.05 significance level with df = 7

P value = .0818

No,because p value is greater than significance level

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