Can somebody classify this figure

Did you get your 8 hours of sleep last night?" is a common question. In a recent survey of 151 postpartum women, the folks at th

Ber [7]

Answer:

There is evidence to suggest that postpartum women do not get enough sleep

Step-by-step explanation:

Given that in a recent survey of 151 postpartum women, the folks at the National Sleep Foundation found that the mean sleep time was 7.8 hours, with a standard deviation of 1.4 hours.

$H_0: \bar x =8\\H_a: \bar x$

(left tailed test at 5% significance)

Mean difference = -0.20

Std error of sample = $\frac{s}{\sqrt{n} } \\=\frac{1.4}{\sqrt{151} } \\=0.114$

Test statistic t = mean difference/std error =

= -1.76

df = 150

p value one tailed = 0.0402

Since p < alpha, we reject H0

There is evidence to suggest that postpartum women do not get enough sleep

8 0

1 year ago

How much oil wells in a given field will ultimately produce is key information in deciding whether to drill more wells. Data was

Ugo [173]

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

3 0

2 years ago

The Greens want to put an addition on their house 18 months from now. They will need to save $10,620 in order to achieve this go

ValentinkaMS [17]

Option A. is the correct answer :)

4 0

1 year ago

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past

Volgvan

Answer: We reject the null hypothesis, and we use Normal distribution for the test.

Step-by-step explanation:

Since we have given that

We claim that

Null hypothesis : $H_0:\mu\geq 50000$

Alternate hypothesis : $H_1:\mu$

There is 5% level of significance.

$\bar{X}=46800\\\\\sigma=9800\\\\n=29$

So, the test statistic would be

$z=\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{46800-50000}{\dfrac{9800}{\sqrt{29}}}\\\\z=-1.75$

Since alternate hypothesis is left tailed test.

So, p-value = P(z≤-2.31)=0.0401

And the P-value =0.0401 is less than the given level of significance i.e. 5% 0.05.

So, we reject the null hypothesis, and we use Normal distribution for the test.

4 0

1 year ago

The probability that a person in the United States has type B+ blood is 12%. Three unrelated people in the United States are s

V125BC [204]

Answer:

The probability that all three have type B+ blood is 0.001728

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have type B+ blood, or they do not. The probability of a person having type B+ blood is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$