Math / Data & Statistics

Question```latex \text{Homework } \mathbf{f} \, 5: 9(1,3,4,5) \, 14(1,2) \\ \text{Question 10 of 30 (1 point) I Question Attempt: 1 of 3} \\ \text{Jonathan} \\ 6 \\ 7 \\ 8 \\ \checkmark 9 \\ 10 \\ 11 \\ 12 \\ 13 \\ 14 \\ 15 \\ 16 \\ \text{Españ} \\ 17 \\
\text{Spam: A researcher reported that } 71.8\% \text{ of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be } 77\%. \text{ He examines a random sample of 500 emails received at an email server, and finds that 364 of the messages are spam. Can you conclude that the percentage of emails that are spam differs from } 77\% \text{? Use both } \alpha=0.01 \text{ and } \alpha=0.05 \text{ levels of significance and the } P\text{-value method with the TI-84 Plus calculator.}
\text{Part 1 of 5} \\ \text{(a) State the appropriate null and alternate hypotheses.} \\ H_{0}: \square^{\infty} \\
\text{Part 3 of 5} \\ \text{(c) Compute the } P\text{-value. Round the answer to at least four decimal places.} \\ P\text{-value }=0.05^{\otimes} \\
\text{Correct Answer:} \\ P\text{-value }=0.0256 \\
\text{Part 4 of 5} \\ \text{(d) Determine whether to reject } H_{0}. \\
\text{At the } \alpha=0.01 \text{ level, do not reject } \nabla \text{ the null hypothesis } H_{0}. \\ \text{At the } \alpha=0.05 \text{ level, reject } \quad \boldsymbol{r} \text{ the null hypothesis } H_{0}. \\
\text{Part: } 4 / 5 \\
\text{Part 5 of 5} \\ \text{(e) State a conclusion.} \\
\text{At the } \alpha=0.01 \text{ level of significance, there } \square \text{ enough evidence to conclude that the percentage of emails that are spam differs from } 77\%. \\ \text{At the } \alpha=0.05 \text{ level of significance, there } \square \text{ (Choose one) enough evidence to conclude that the percentage of emails that are spam differs from } 77\%. \\ \text{Skip Part} \\ \text{Check} \\ \text{Save For Later} \\ \text{Submit Assignment}

Studdy Solution

STEP 1

1. We are testing whether the percentage of spam emails differs from 77%.
2. A sample of 500 emails shows 364 are spam.
3. We use the P-value method with significance levels $\alpha = 0.01$ and $\alpha = 0.05$ .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Compute the P-value.
4. Determine whether to reject the null hypothesis.
5. State the conclusion.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The percentage of spam emails is 77%, i.e., $p = 0.77$ . - Alternative hypothesis ( $H_a$ ): The percentage of spam emails is not 77%, i.e., $p \neq 0.77$ .

STEP 4

Calculate the test statistic using the formula for a proportion:
$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$
Where: - $\hat{p} = \frac{364}{500} = 0.728$ - $p_0 = 0.77$ - $n = 500$
Calculate:
$z = \frac{0.728 - 0.77}{\sqrt{\frac{0.77 \times 0.23}{500}}}$

STEP 5

Compute the P-value using the calculated $z$ -value. The P-value is the probability that the observed data would occur if the null hypothesis were true. Use a standard normal distribution table or calculator.
Given the correct P-value is $0.0256$ .

STEP 6

Determine whether to reject the null hypothesis:
- At $\alpha = 0.01$ : Since $0.0256 > 0.01$ , do not reject $H_0$ . - At $\alpha = 0.05$ : Since $0.0256 < 0.05$ , reject $H_0$ .

STEP 7

State the conclusion:
- At the $\alpha = 0.01$ level of significance, there is not enough evidence to conclude that the percentage of emails that are spam differs from 77%. - At the $\alpha = 0.05$ level of significance, there is enough evidence to conclude that the percentage of emails that are spam differs from 77%.

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Studdy Solution

STEP 1

1. We are testing whether the percentage of spam emails differs from 77%.
2. A sample of 500 emails shows 364 are spam.
3. We use the P-value method with significance levels $\alpha = 0.01$ and $\alpha = 0.05$ .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Compute the P-value.
4. Determine whether to reject the null hypothesis.
5. State the conclusion.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The percentage of spam emails is 77%, i.e., $p = 0.77$ . - Alternative hypothesis ( $H_a$ ): The percentage of spam emails is not 77%, i.e., $p \neq 0.77$ .

STEP 4

Calculate the test statistic using the formula for a proportion:
$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$
Where: - $\hat{p} = \frac{364}{500} = 0.728$ - $p_0 = 0.77$ - $n = 500$
Calculate:
$z = \frac{0.728 - 0.77}{\sqrt{\frac{0.77 \times 0.23}{500}}}$

STEP 5

Compute the P-value using the calculated $z$ -value. The P-value is the probability that the observed data would occur if the null hypothesis were true. Use a standard normal distribution table or calculator.
Given the correct P-value is $0.0256$ .

STEP 6

Determine whether to reject the null hypothesis:
- At $\alpha = 0.01$ : Since $0.0256 > 0.01$ , do not reject $H_0$ . - At $\alpha = 0.05$ : Since $0.0256 < 0.05$ , reject $H_0$ .

STEP 7

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Studdy Solution

STEP 1

1. We are testing whether the percentage of spam emails differs from 77%.2. A sample of 500 emails shows 364 are spam.3. We use the P-value method with significance levels α=0.01\alpha = 0.01α=0.01 and α=0.05\alpha = 0.05α=0.05.

STEP 2

1. State the null and alternative hypotheses.2. Calculate the test statistic.3. Compute the P-value.4. Determine whether to reject the null hypothesis.5. State the conclusion.

STEP 3

State the null and alternative hypotheses:- Null hypothesis (H0H_0H0​): The percentage of spam emails is 77%, i.e., p=0.77p = 0.77p=0.77. - Alternative hypothesis (HaH_aHa​): The percentage of spam emails is not 77%, i.e., p≠0.77p \neq 0.77p=0.77.

STEP 4

STEP 5

Compute the P-value using the calculated zzz-value. The P-value is the probability that the observed data would occur if the null hypothesis were true. Use a standard normal distribution table or calculator.Given the correct P-value is 0.02560.02560.0256.

STEP 6

Determine whether to reject the null hypothesis:- At α=0.01\alpha = 0.01α=0.01: Since 0.0256>0.010.0256 > 0.010.0256>0.01, do not reject H0H_0H0​. - At α=0.05\alpha = 0.05α=0.05: Since 0.0256<0.050.0256 < 0.050.0256<0.05, reject H0H_0H0​.

STEP 7

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1. We are testing whether the percentage of spam emails differs from 77%.
2. A sample of 500 emails shows 364 are spam.
3. We use the P-value method with significance levels $\alpha = 0.01$ and $\alpha = 0.05$ .

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Compute the P-value.
4. Determine whether to reject the null hypothesis.
5. State the conclusion.

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The percentage of spam emails is 77%, i.e., $p = 0.77$ . - Alternative hypothesis ( $H_a$ ): The percentage of spam emails is not 77%, i.e., $p \neq 0.77$ .

Compute the P-value using the calculated $z$ -value. The P-value is the probability that the observed data would occur if the null hypothesis were true. Use a standard normal distribution table or calculator.
Given the correct P-value is $0.0256$ .

Determine whether to reject the null hypothesis:
- At $\alpha = 0.01$ : Since $0.0256 > 0.01$ , do not reject $H_0$ . - At $\alpha = 0.05$ : Since $0.0256 < 0.05$ , reject $H_0$ .