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Traffic police monitor the speed of vehicles as they travel over a new bridge. The average speed for a sample of 41 vehicles was 81.35 km/h, with the sample standard deviation being 4.52km/h. We will assume that the speeds are Normally distributed, and the police are interested in the mean speed.
Part a) Since the variance of the underlying Normal distribution is not known, inference here would involve the t distribution. How many degrees of freedom would the relevant t distribution have?
Part b) Create a 95 % confidence interval for the mean speed of vehicles crossing the bridge. Give the upper and lower bounds to your interval, each to 2 decimal places. ( ,)
Part c) The police hypothesized that the mean speed of vehicles over the bridge would be the speed limit, 80 km/h. Taking a significance level of 5 %, what should infer about this hypothesis?
A. We should reject the hypothesis since 80 km/h is in the interval found in (b).
B. We should reject the hypothesis since 80 km/h is not in the interval found in (b).
C. We should not reject the hypothesis since the sample mean is in the interval found in (b).
D. We should reject the hypothesis since the sample mean was not 80 km/h.
E. We should not reject the hypothesis since 80 km/h is in the interval found in (b).
Part d) Decreasing the significance level of the hypothesis test above would (select all that apply)
A. not change the Type II error probability.
B. decrease the Type I error probability.
C. either increase or decrease the Type I error probability.
D. not change the Type I error probability.
E. increase the Type I error probability.

Respuesta :

Answer:

(a) The degrees of freedom; the relevant t distribution would have = n - 1 = 41 - 1 = 40.

(b) A 95% confidence interval for the mean speed of vehicles crossing the bridge is [79.92 km/hr, 82.78 km/hr] .

(c) We should not reject the hypothesis since 80 km/h is in the interval found in (b).

(d) Decrease the Type I error probability.

Step-by-step explanation:

We are given that the average speed for a sample of 41 vehicles was 81.35 km/h, with the sample standard deviation being 4.52km/h.

(a) Since here we don't know about population standard deviation, so the distribution that will be used here is t-distribution as the data also follows the normal distribution.

The degrees of freedom; the relevant t distribution would have = n - 1 = 41 - 1 = 40.

(b) Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                             P.Q.  =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~   [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample average speed of vehichles = 81.35 km/h

             s = sample standard deviation = 4.52 km/h

             n = sample of vehicles = 541

            [tex]\mu[/tex] = population mean  

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;

P(-2.021 < [tex]t_4_0[/tex] < 2.021) = 0.95  {As the critical value of t at 40 degrees of

                                               freedom are -2.021 & 2.021 with P = 2.5%}  

P(-2.021 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.021) = 0.95

P( [tex]-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.021 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.021 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.021 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.021 \times {\frac{s}{\sqrt{n} } }[/tex] ]

                                      = [ [tex]81.35-2.021 \times {\frac{4.52}{\sqrt{41} } }[/tex] , [tex]81.35+2.021 \times {\frac{4.52}{\sqrt{41} } }[/tex] ]  

                                     = [79.92 km/hr, 82.78 km/hr]

Therefore, a 95% confidence interval for the mean speed of vehicles crossing the bridge is [79.92 km/hr, 82.78 km/hr] .

(c) The hypothesis given to us is;

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 80 km/hr    {means that the mean speed of vehicles over the bridge would be the speed limit of 80 km/h}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] [tex]\neq[/tex] 80 km/hr    {means that the mean speed of vehicles over the bridge would be the speed limit of different from 80 km/h}

Here, we should not reject the null hypothesis since 80 km/h is in the interval found in (b) which means that we are 95% confident that the mean speed of vehicles over the bridge would be the speed limit of 80 km/h.

(d) Decreasing the significance level of the hypothesis test above would decrease the Type I error probability because Type I error probability is represented by the significance level.

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