A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. Of the 29 tires surveyed, the mean lifespan was 45,800 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly consistent with the claim?

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ ≥ 50000

For the alternative hypothesis,

µ < 50000

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = lifetime of the tyres

µ = mean lifetime

σ = standard deviation

n = number of samples

From the information given,

µ = 50000 miles

x = 45800 miles

σ = 8000

n = 29

z = (50000 - 45800)/(8000/√29) = - 2.83

Looking at the normal distribution table, the probability corresponding to the z score is 0.9977

Since alpha, 0.05 < than the p value, 0.9977, then we would accept the null hypothesis. Therefore, At a 5% level of significance, the data is not highly consistent with the claim.

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

hello 0xxbadwolfxx0,

2.6 x 10^9 is 2,600,000,000.

1.3 x 10^2 is 130

multiplying 130 and 2,600,000,000 would simply give you 200,000,000,000.

therefore, the answer would be d. 2.0 × 10^11.

i hope i ! warm regards, mae.