Question

1. Tony claims that compared to other students in his school he saves an average of $7.50 per month on public transit, because he understands all the available ways to use transit transfers. His friend does not believe this claim and collects a random sample of Tony's monthly spending (he takes public transit 22 days in a month). The sample shows that Tony saves an average of $6.45 compared to other students (with a sample standard deviation of $0.89).
   1. Construct a 99% confidence interval for Tony’s savings.
   2. Can Tony's claim be refuted at the 1% level of significance? Explain.

Answer 1

Answer)

As the population s.d is unknown here we will use t distribution to estimate the answer

Degrees of freedom is = n-1 = 22 -1 = 21

Critical value t from t table for 21 degrees of freedom and 99% confidence level is = 2.83

Margin of error (MOE) = t*s.d/√n = 2.83*0.89/√22 = 0.5372

Confidence interval is given by

Mean - MOE < u < Mean + MOE

Mean = 6.45

5.9128 < u < 6.9872

B)

Null hypothesis Ho : u = 7.5

Alternate hypothesis Ha : u not equal to 7.5

As the interval does not contain the null hypothesised value 7.5

We reject the null hypothesis

We do not have enough evidence to support the claim