Question

The mean work week for engineers in start-up companies is claimed to be about 64 hours with a standard deviation of 8 hours. Kara, a newly hired engineer hopes that it's not (she wants it to be shorter). She asks 10 engineers in start-ups for the lengths of their mean work weeks. Their responses are 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.
1. Is the newly hired engineer testing a mean or a proportion?
a. proportion.
b. mean.
2. The null hypothesis (or company claim), in symbols, is mu =___.
3. If the claim is correct, what is the probability that a sample mean would be as low as Kara's mean?
4. If Kara will accept the 60 hours as the real average so long as there's at least a 10% chance the sample average value could be as low as hers, should she accept the claim?
a. yes.
b. no.

Answer 1

Answer:

1.) mean

2.) H0 : μ = 64

3.) 0.0028

4) Yes

Step-by-step explanation:

Null hypothesis ; H0 : μ = 64

Alternative hypothesis ; H1 : μ < 64

From the data Given :

70; 45; 55; 60; 65; 55; 55; 60; 50; 55

Using calculator :

Xbar = 57

Sample size, n = 10

Standard deviation, s = 7.14

Test statistic :

(xbar - μ) ÷ s/sqrt(n)

(57 - 64) ÷ 8 / sqrt(10)

Test statistic = - 2.77

Pvalue = (Z < - 2.77) = 0.0028 ( Z probability calculator)

α = 10% = 0.1

Reject H0 ; if P < α

Here,

P < α ; Hence, we reject the null