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Given In the question that the company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs that they make is 3 pounds. One of the chairs collapsed beneath a $220$-pound student last week. You wonder whether the manufacturer is exaggerating the breaking strength of the chairs. we have to State null and alternative hypotheses in words and symbol. we have to State null and alternative hypotheses in words and symbols.

For high school students, a manufacturer makes classroom chairs. The chairs' average breaking strength is $300$pounds. One of the chairs, weighing $220$pounds, collapsed.

Let $\mathrm{μ}$represent the chair's strength.

The null hypothesis is the one that is thought to be correct. $H_0$is the symbol for it.

Determine what the null hypothesis is. $H_0:\mathrm{μ}=300$

One of the seats, weighing less than$300$pounds, gave apart.

Create an alternative hypothesis.$H_1:\mathrm{μ}<300$.

Given In the question that the company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs that they make is $300$pounds. One of the chairs collapsed beneath a $220$-pound student last week. You wonder whether the manufacturer is exaggerating the breaking strength of the chairs. we have to State null and alternative hypotheses in words and symbol. we have to describe a Type I error and a Type II error in this situation, and give the consequences of each.

When $H_0$is rejected, it is referred to as a Type I error. $H_0:\mathrm{μ}=300$is the null hypothesis.

The chair's mean breaking strength is $300$pounds.

The Type Il error occurs when $H_0$is accepted when $H_0$is false.

$H_1:\mathrm{μ}<300$is an alternative hypothesis.

This means the chair's mean breaking strength is less than $300$pounds.

Type II errors in the provided problem imply that collapsing a chair is unlikely.

The type Il mistake indicates that the likelihood of the chair collapsing is greater.

Given in the question that there are $3$significane levels $.01,0.05,0.10$. we have to find out that which one is recommended for the test.

The three levels of significance are$0.01,0.05and0.10$, respectively.

When $H_0$is true and $H_0$is rejected, it is referred to as a Type I error.

$H_0:\mathrm{μ}=300$is the null hypothesis.

When $H_0$is accepted when $H_0$is false, this is known as a Type II error.

$H_1:\mathrm{μ}<300$is an alternat ve hypothesis.

Type I errors imply that the chair will collapse less in response to the provided situation.

The Type Il fault means the chair will collapse more frequently.

The major goal is to keep Type II errors to a minimum

because they are more dangerous.

This occurs when the Type I mistake $\mathrm{Î\pm }$is higher.

The value also de$\mathrm{Î\pm }$notes the significance level.

$0.01,0.02,0.10$are the specified values.

We can choose significance level $0.10$, because this is the greatest value out of the given values .

Given in the question that the power of this test to detect $\mathrm{μ}=294$is $0.71$we have to explain what this means to someone who knows little statistics.

$H_0:\mathrm{μ}=300$is the null hypothesis.

$H_1:\mathrm{μ}<300$is an alternative hypothesis.

The test's power to detect $\mathrm{μ}=294is0.71$. Test power is defined as the chance of rejecting the null hypothesis if the alternative hypothesis is true.

When $\mathrm{μ}=294$, the test's power is $0.71$, which means the chance of rejecting the null hypothesis is $0.71$.

We have to explain two ways that you could increase the power of the test from (d).

1. Increase the sample size: A larger sample size allows for more accurate estimation.

2. Increase the level of significance: This will increase the probability of making a Type I error, lowering the chances of making a Type II error. As a result, it will increase the test's power by definition.