Question

You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002.

      Ho:μ=58.2Ho:μ=58.2
      Ha:μ<58.2Ha:μ<58.2

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=24n=24 with mean M=41M=41 and a standard deviation of SD=20.6SD=20.6.

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

2. Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 398 drivers and find that 312 claim to always buckle up. Construct a 93% confidence interval for the population proportion that claim to always buckle up.

Box 1: Enter your answer using interval notation. Example: [2.1,5.6172)
Use U for union to combine intervals. Example: (-oo,2] U [4,oo)
Enter DNE for an empty set. Use oo to enter Infinity.

Answer 1

1) The test statistic t = ()/(s/)

= (41 - 58.2)/(20.6/)

= -4.257

P-value = P(T < -4.257)

= 0.0001

2) = 312/398 = 0.7839

At 93% confidence interval the critical value is z_0.035 = 1.81

The 93% confidence interval is

+/- z_0.035 * sqrt((1 - )/n)

= 0.7839 +/- 1.81 * sqrt(0.7839 * (1 - 0.7839)/398)

= 0.7839 +/- 0.0373

= 0.7466, 0.8212