T confidence interval Introduction complete solutions correct answers key

t confidence interval Introduction complete solutions correct answers key

We are interested in finding the service time of buses or time taken by the bus from one route to another. The service provider company is claiming that it takes 50 minutes to cover this distance. So we are interested in finding whether there claim is right or whether it takes more than three hours to cover that particular distance.

Null and Alternative Hypothesis:

Null Hypothesis (Ho): The population mean travelling time is equal to 50 minutes from one route to another.

Alternative Hypothesis (Ha): The population mean travelling time is more than 50 minutes from one route to another.

Mathematically;

Null hypothesis                          H₀: µ = 50 and

Alternative Hypothesis                Ha: µ > 50

Where µ is the population mean of travelling time.

Level of significance = 5% or 0.05

Test statistics

Here the test statistics is t test for one sample

Decision Rule:

We will reject our null hypothesis if t >t (tabulated value)

t (tabulated value at α = .05 with degree of freedom) = 1.65

Methodology

Data Collection:

We have collected sample of travelling from the same company bus with the same route and we have collected data for vehicle number 7581, so we have 40 observations and the mean of these observation is 52.65 with a standard deviation 0.8. In other way we can say that the sample mean is 52.65 and Standard deviation is .8. Now we are interested in finding whether the claim of company is true that is the average mean time is 50 minutes or it is more than 50 minutes based on the sample information we have collected. 

Calculations:

Here independent variable is route and dependent variable is time taken.

Here we have to use testing procedure

Here, n= 40 and  = .8 (calculated values)

Here α = .05

Test statistics is given by

t =  

Test statistics under H₀ is given by

t =  , where  is the specified value of mean that is 50 here

t =  =    =  = 21.03

Now we have t = 21.03 and we have to compare this value with the table values,

Results

Since 21.03 > 1.685, therefore we will reject our null hypothesis and the alternative is accepted. It means we say that the mean time taken from that particular route is bigger than 50 minutes. So we can say that the mean travelling time is more than 50 minutes for The B63 MTA bus.

Limitations

These findings are applicable to the data collected and analyzed. We cannot generalize these conclusion and finding.

I just need the following thing ,my instructor want me to do :-

You did not provide the 95% confidence interval. The test statistic calculation is wrong. Rewrite your conclusion in non-technical terms, See page 391 in your textbook.