A manufacturer claims that its rechargeable batteries are good for an average of more than 1.000 charges. A random sample of 100 batteries has a mean life of 1002 charges and a standard deviation of 14. Is there enough evidence to support this claim at a significance level of 0.01? a. State the hypotheses.
b. State the test statistie information
c. State either the p-value or the critical information d. State your conclusion and explain your reasoning

It's 1000 charges and not 1.000 charges

A)Null Hypothesis;H0: μ = 1000

Alternative Hypothesis;Ha: μ ≠ 1000

B) t-statistic = 1.4286

C) p-value = 0.15628

D) We conclude that we will fail to reject the manufacturers claim that its rechargeable batteries are good for an average of more than 1000 charges

Step-by-step explanation:

We are given;

x = 1002 charges

s = 14

μ = 1000 charges

n = 100

degree of freedom = n - 1 = 100 - 1 = 99

A) The hypotheses are;

Null Hypothesis;H0: μ = 1000

Alternative Hypothesis;Ha: μ ≠ 1000

B) t-statistic = (x - μ)/(s/√n)

(1002 - 1000)/(14/√100) = 1.4286

C) From the t-score calculator results attached, the p-value is approximately 0.15628

D) The P-value of 0.15628 is is greater than the significance level of 0.01, thus we fail to reject the null hypothesis, and we conclude that the result is statistically nonsignificant.

b. x- 48 =120

step-by-step explanation:

in the math equation 8 x 7 = 56, the 8 and the 7 are both factors of 56, since in this lesson we will use square tiles to figure out the factors of a whole find as many different arrays as you can, but follow one important rule: all the rows must have the i can make two rows of nine, so 2 and 9 are both factors of 18.

step-by-step explanation: