Answer:
<h2>5 + 3w + 3 - w = 2w + 8</h2>Step-by-step explanation:
PLEASE Please PLEASE HELP LOTS OF POINTS!!!!!! What is the slope and Y-intercept of this graph?
2 answers:
You might be interested in
Answer:
We conclude that a compact microwave oven consumes a mean of more than 250 W.
Step-by-step explanation:
We are given that an appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W with a population standard deviation of 15 W.
They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W.
Let = <u><em>mean power consumption for microwave ovens.</em></u>
So, Null Hypothesis, :
250 W {means that a compact microwave oven consumes a mean of no more than 250 W}
Alternate Hypothesis, :
> 250 W {means that a compact microwave oven consumes a mean of more than 250 W}
The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean power consumption for ovens = 257.3 W
σ = population standard deviation = 15 W
n = sample of microwave ovens = 20
So, <em><u>the test statistics</u></em> =
= 2.176
The value of z test statistics is 2.176.
<u>Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.</u>
Since our test statistic is more than the critical value of t as 2.176 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.
Therefore, we conclude that a compact microwave oven consumes a mean of more than 250 W.
Answer:
The mean is 39
The median is 42
Step-by-step explanation:
The median is the best measure of center for the above data because half of the values are less than the median and half of the values are more than the median. It's probably the best measure of center to use in a skewed distribution.
The measure of variability is the range because is the simplest measure of variability to calculate but can be misleading if the dataset contains extreme values.
The variability is 67.25
This is how I did it
