A company wants to establish that the mean life of its batteries, when used in a wireless mouse, is over 183 days. The data will
1 answer:
Answer:
a) Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
b) If the true mean is 190 days, Type II error can be made.
Step-by-step explanation:
Let mu be the mean life of the batteries of the company when it is used in a wireless mouse
Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
Type II error happens if we fail to reject the null hypothesis, when actually the alternative hypothesis is true.
That is if we conclude that mean life of the batteries of the company when it is used in a wireless mouse is at most 183 days, but actually mean life is 190 hours, we make a Type II error.
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Answer:
1.7 × 10⁻⁴
Step-by-step explanation:
The question relates to a two sample z-test for the comparison between the means of the two samples
The null hypothesis is H₀: μ₁ ≤ μ₂
The alternative hypothesis is Hₐ: μ₁ > μ₂
Where;
= 13.5
= 12
σ₁ = 2.5
σ₂ = 1.5
We set our α level at 0.05
Therefore, our critical z = ± 1.96
For n₁ = n₂ = 23, we have;
We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96
Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983
Therefore;
The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.
This one bc there’s one point vertically and there’s no more than one point when you draw a vertical line down
