A: The middle 95% contain between 23.8 and 24.2 oz.(2 SDs under and over the mean) B: Approximately 68% of boxes have between 23.9 and 24.1 oz of cereal (1 SD under and over the mean) C: 2.5% of boxes contain more than 24.2, because 95% are within 2 SDs, and that includes lower and higher extremes. D: 16% of boxes have more than 24.1, because 68% are within 1 SD of the mean, so 32 are left, and that includes lower and higher extremes.

3 0

3 years ago

Find the common difference for the given sequence shown. 56,49,42,35

ddd [48]

Answer:

-7

Step-by-step explanation:

let the first term be a

thus, a=56

common difference d= U2-U1 = 49-56= -7

4 0

3 years ago

In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer

stira [4]

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = percentage of patients stop smoking when given sustained care.

So, Null Hypothesis, $H_0$ : p = 80% {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, $H_A$ : p $\neq$ 80% {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of patients who stop smoking when given sustained care = 80.2%

n = sample of patients = 192

So, the test statistics = $\frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }$

= 0.07

The value of z test statistics is 0.07.

Also, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.07) = 1 - P(Z $\leq$ 0.07)

= 1 - 0.52790 = 0.4721

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.

8 0

3 years ago

Helpppppppppp asappppppppp 20 points

Lesechka [4]

Here’s a small hint: “miles per hour”

Put the amount of miles over the amount of hours then calculate the amount from there! It’s either the 2nd option or the 4th one! Hope it’s helpful!

4 0

2 years ago

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