Im giving you 30 points just help me :((((((((

A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher a

erica [24]

Answer:

z=-1.591

Step-by-step explanation:

Null Hypotheses,

$H_0 : p=0.8\\H_a: p>0.80$

So we use z-test for one population proportion (right-tailed test)

According to this informaton, we defined that

$\alpha=0.05$

$z_c=1.64$ (critical value)

So our Rejection region is $R={z:z>1.64}$

$z=\frac{\bar{p}-p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.755-0.8}{\sqrt{0.8(1-0.8)/200}}=-1.591$

Not rejection.

6 0

3 years ago

The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the

lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

Let $\mu$ = average sodium content.

So, Null Hypothesis, $H_0$ : $\mu$ = 1100 milligrams {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 1100 milligrams {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample average sodium content = 1088.64 milligrams

s = sample standard deviation = 234.12 milligrams

n = sample of packages of Oriental Spice Sauce = 40

So, test statistics = $\frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}$ ~ $t_3_9$

= -0.307

The value of z test statistics is -0.307.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, 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 the sodium content is same as what the nutrition label states.

3 0

4 years ago

Which fraction goes in the blank to make the comparison true? __________>1/3

Oksana_A [137]

Answer:

1/2, 2/3, 3/4, etc

i think thats what you need

Step-by-step explanation:

8 0

3 years ago

MAKING BRAINLIEST!!Bruce Wayne splits his budget by using the ratio 12:7. For every $12 he spends

Anvisha [2.4K]

D. 28

this is because 48 divided by 12 is 4, 4 times 7 is 28.

3 0

3 years ago

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My bed is 3 inches by 2and 2/3 inches by 1/3 inch. What is the volume of my bed?

ki77a [65]

Step-by-step explanation:

$volume \: of \: bed \\ \\ = 3 \times 2 \frac{2}{3} \times \frac{1}{3} \\ \\ = 3 \times \frac{8}{3} \times \frac{1}{3} \\ \\ = \frac{8}{3} \\ \\ = 2 \frac{2}{3} \: {inch}^{3}$

7 0

3 years ago

Read 2 more answers