Question

1. A police officer is concerned about speeds on a certain section of Interstate 95. The data accompanying this exercise show the speeds of 40 cars on a Saturday afternoon.

a. The speed limit on this portion of Interstate 95 is 66 mph. Specify the competing hypotheses in order to determine if the average speed is greater than the speed limit.

• H₀: μ = 66; H_A: μ ≠ 66

• H₀: μ ≥ 66; H_A: μ < 66

• H₀: μ ≤ 66; H_A: μ > 66

b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

b-2. Find the p-value.

• 0.025 p-value < 0.05
• 0.01 p-value < 0.025

• p-value < 0.01

• p-value 0.10
• 0.05 p-value < 0.10

c. At α = 0.05, are the officer’s concerns warranted?

2. Consider the following hypotheses:

H₀: μ = 8,900
H_A: μ ≠ 8,900

The population is normally distributed with a population standard deviation of 800. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)

		Test statistic	p-value
a.	x−x− = 8,940; n = 110
b.	x−x− = 8,940; n = 260
c.	x−x− = 8,680; n = 38
d.	x−x− = 8,710; n = 38

Answer 1

2) answer)

Z(test statistics) = (obtained mean - claimed mean)/(standard deviation/√n)

A)

Obtained mean = 8940

And claimed = 8900

Standard deviation = 800

And n = 110

Substituting these values we get

Z = (8940-8900)/(800/√110)

Z = 0.52

And from z table, 0.52 corresponds to 0.6985

So p-value = 2*(1-0.6985)

As this is two tailed test because there is no particular direction so we need to find the area of greater than 0.52 then this will be the area of one tail

So we will multiply it with 2 to get the p-value for two tails

Test statistics = 0.52

P-value = 0.6030

As the p-value is greater than 0.1(10% significance level)

We fail to reject the null hypothesis

B)

Here every thing is same just n is =260

Z =(8940-8900)/(800/√260)

Test statistic = 0.81

From z table, 0.81, corresponds to 0.7910

Therefore, p-value is = 2*(1-7910) = 0.4180

Here also p-value is greater than 0.1, we fail to reject null hypothesis.

C)

Obtained = 8680

Claimed = 8900

N=38

Z =(8680-8900)/(800/√38)

Z = −1.6952138508

Test statistics = -1.70

From z table, -1.7 corresponds to 0.0446

Now here as we know z table shows the value less than certain value

So here we are already given with one tail (less than -1.7)

We just need to multiply it with 2, to get the required p-value for two tails

P-value = 0.0892

As p-value is less than 0.1, we reject the null hypothesis

D)

Z =(8710-8900)/(800/√38)

Test statistics = -1.46

P-value = 2*0.0721 = 0.1442

P-value is greater than 0.1, we fail to reject the hypothesis.