Question

-A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every bag contains exactly the same weight. But if more than 15% of bags are over-filled then they stop production to fix the machine.

They define over-filled to be more than 1 ounce above the weight on the package. The engineer weighs 100 bags and finds that 21 of them are over-filled.

He plans to test the hypotheses: H₀: p = 0.15 versus H_a: p > 0.15 (where p is the true proportion of overfilled bags).

What is the test statistic?

1. Z = 1.68
2. Z = -1.68
3. Z = 4
4. Z = -1.47

-According to a Pew Research Center, in May 2011, 35% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students.

She selects 300 community college students at random and finds that 120 of them have a smart phone. In testing the hypotheses: H₀: p = 0.35 versus H_a: p > 0.35, she calculates the test statistic as Z = 1.82.

Use the Normal Table to help answer the p-value part of this question.

Click here to access the normal table.

1. There is enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).
2. There is enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.068).
3. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.966).
4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).

Answer 1

Concepts and reason

Statistical hypotheses testing: Hypotheses testing is used to make inferences about the population based on the sample data. The hypotheses test consists of null hypothesis and alternative hypothesis.

Null hypothesis: The null hypothesis states that there is no difference in the test, which is denoted by ${H_0}$. Moreover, the sign of null hypothesis is equal$\left( = \right)$, greater than or equal $\left( \ge \right)$and less than or equal$\left( \le \right)$.

Alternative hypothesis: The hypothesis that differs from the ${H_0}$ is called alternative hypothesis. This signifies that there is a significant difference in the test. The sign of alternative hypothesis is less than $\left( < \right)$, greater than $\left( > \right)$, or not equal $\left( \ne \right)$.

Z-statistic for proportion:

The standardized z-statistic is defined as the ratio of the ‘distance between observed statistic and the proportion from the null distribution’ and the ‘standard deviation of the proportion’.

P-value:

The probability of getting the value of the statistic that is as extreme as the observed statistic when the null hypothesis is true is called as P-value.

Fundamentals

One sample proportion:

The formula for test statistic of sample proportion is given below:

$z = \frac{{\hat p - p}}{{\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }}$

Where p is the population proportion and $\hat p$is the sample proportion.

Z Test for one sample proportion.

Excel add-in (PHStat) procedure:

1.In EXCEL, Select Add-Ins > PHStat > One-Sample Tests > Z Test for the proportion.

2.Enter ---- under Null hypothesis.

3.Enter ---- under Level of significance.

4.Enter ---- under Number of Items Interest.

5.Enter ---- under Sample size.

6.In Test Options, choose Two (or Upper or Lower) Tail Test.

7.In Output Options, enter a Title and click OK.

Rejection rule for p-value method:

If $p{\rm{ - value}} \le \alpha \left( { = 0.05} \right)$, then reject the null hypothesis.

The Hypotheses for this test is determined below:

From the given information, the random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The level of significance is 0.05.

Null hypothesis:

${H_0}:p = 0.90$

Alternative Hypothesis:

${H_a}:p < 0.90$

The test statistics is obtained below:

From the given information,$n = 100$,$x = 83$, $p = 0.90$and $\alpha = 0.05$

Instructions to obtain the -value:

1.In EXCEL, Select Add-Ins > PHStat > One-Sample Tests > Z Test for the proportion.

2.Enter 0.15 under Null hypothesis.

3.Enter 0.05 under Level of significance.

4.Enter 21 under Number of Items Interest.

5.Enter 100 under Sample size.

6.In Test Options, choose Upper Tail Test.

7.In Output Options, enter a Title and click OK.

Follow the above instructions to find the following output.

From the output, the test statistics is 1.68.

The P-value is obtained as shown below:

From the given information, $Z = 1.82$.

The required p-value is,

From the “standard normal table”, the area to the left of $Z = 1.82$ is,

Procedure for finding the probability value (confidence level) from standard normal table:

The z value be 1.82.

1.In the standard normal table first locate the value 1.82 in the first z column.

2.Locate the value of 1.80 in the first z row.

3.Move right until the column of 0.02 is reached.

4.From the located 0.02 column move down until the row 1.82 is reached.

5.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

The area to the left of $Z = 1.82$ is 0.9656.

$\begin{array}{c}\\P\left( {Z > 1.82} \right) = 1 - P\left( {Z < 1.82} \right)\\\\ = 1 - 0.9656\\\\ = 0.0344\\\end{array}$

The conclusion is stated below:

The p-value is 0.034 and the level of significance is 0.05.

The p-value is less than level of significance.

That is, $p{\rm{ - value}}\left( { = 0.034} \right) < \alpha \left( { = 0.05} \right)$

By the rejection rule, reject the null hypothesis.

Therefore, it can be concluded that there is an enough to show that more than 35% of community college students own a smart phone (P-value = 0.034)

Ans:

The test statistics is $z = 1.68$.

There is enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).