You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,298; $1,229.96; $1,193.07; $1,279; $1,224.98; $1,229.97; $1,269.98; $1,248.
Can you argue that pricing has a larger standard deviation than claimed by the manufacturer?

We fail to reject the null

Step-by-step explanation:

Given the data: 1298; 1229.96; 1193.07; 1279; 1224.98; 1229.97; 1269.98; 1248

Mean price, μ = 1249

Standard deviation, σ = 25

Variance, σ² =

Null hypothesis, H0 : σ² = 25²

Alternative hypothesis, H1 : σ² > 25

Using calculator, the sample standard deviation :

Sample standard deviation = s = 34.072

Sample size, n = 8

For chi² distribution :

X² = [(n - 1)s²] ÷ σ²

X² = [(8 - 1)*34.072^2] ÷ 25^2

X² = (7 * 1160.901184) / 625

X² = 13.002093

Using the Pvalue from Chisquare calculator, df = 8 - 1 = 7; α = 0.05

Pvalue =0.07206

0.07206 > 0.05

Since , Pvalue > α

We fail to reject the null; there is insufficient evidence to conclude that standard devuatue do selling a computer is greater than 25

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.

its a, i had the same answer, its from study island right

step-by-step explanation: