Is 0.48 greater than 6/15

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin

Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

Let $\mu$ = true average writing lifetime under controlled conditions

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}

The test statistics that is used here is one-sample t test statistics;

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

where, $\bar X$ = sample mean

s = sample standard deviation

n = sample size of pens = 18

n - 1 = degree of freedom = 18 -1 = 17

Now, the decision rule based on the critical value of t is given by;

If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then we will reject our null hypothesis as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.

8 0

3 years ago

1. A square pizza box has an area of 155 cm^2. Which is the best estimate of the side length of the pizza box?

natulia [17]

Answer:

its D

Step-by-step explanation:

did it

7 0

3 years ago

Read 2 more answers

In March 2015, a Nielsen global online survey "found that consumers are increasingly willing to pay more for socially responsibl

Veseljchak [2.6K]

Answer:

For this case the parameter of interest is given by:

$p$ who represent the true proportion of respondents said that they were willing to pay more for products and services from companies who are committed to positive social and environmental impact

For this case we have an estimation given for this parameter. The estimation comes from a sample of 30000 people selected in 60 countries and they got:

$\hat p = 0.66$

This value represent the best estimator for the true proportion since is an unbiased estimator of the real parameter:

$E(\hat p) = p$

Step-by-step explanation:

For this case the parameter of interest is given by:

$p$ who represent the true proportion of respondents said that they were willing to pay more for products and services from companies who are committed to positive social and environmental impact

For this case we have an estimation given for this parameter. The estimation comes from a sample of 30000 people selected in 60 countries and they got:

$\hat p = 0.66$

This value represent the best estimator for the true proportion since is an unbiased estimator of the real parameter:

$E(\hat p) = p$

For this case if we want to test if the population proportion is equal to an specified value we can use the one sample z test for a proportion:

Null hypothesis: $p=p_0$

Alternative hypothesis: $p \neq p_o$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .