Stats homework please help

Question

1. Using a t-distribution, find the margin of error given 𝑐 = .99, 𝑠 = 3, 𝑛 = 6

(a) 4.54 (b) 4.94 (c) 1.21 (d) 12.16

2. Construct a 92% confidence interval for the following sample assuming the population standard deviation is $85.50. A random sample of 34 home theater systems has a mean price of $452.80.

(a) (428.7,476.9) (b) (426.3,479.3) (c) (59.8,111.2) (d) (427.1,478.5)

3. A pit crew claims that its mean pit stop time is less than 13 seconds. Interpret a decision that rejects the null hypothesis.

(a) There is enough evidence to reject the claim that its mean pit stop time is less than 13 seconds.

(b) There is not enough evidence to reject the claim that its mean pit stop time is less than 13 seconds.

(c) There is enough evidence to support the claim that its mean pit stop time is less than 13 seconds.

(d) There is not enough evidence to support the claim that its mean pit stop time is less than 13 seconds.

Answer 1

1. Margin of error = tcritical * s/n^0.5, where t-critical is t score for alpha = (1-0.99)/2 = 0.005 and n-1=5 degrees of freedom. t-critical = 4.032, so margin of error = 4.032 * 3/6^0.5 =~ 4.94 (b)
2. CI = mean +/- tcritical * s/n^0.5, where t-critical is t score for alpha = (1-0.92)/2 = 0.04 and n-1=33 degrees of freedom. t-cricitcal = 1.806. CI = 452.8 +/- 1.806* 85.50/34^0.5 = 452.8 +/- 26.48 = (426.32,479.28) ~= (426.3, 479.3) (b)
3. a)