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3 years ago

The highest grade on a quiz was 95 the lowest was 72. write an inequality to show the range.

1 answer:

7 0

Since the minimum is 72 the every grade must be equal or bigger than it
also 95 is the maximum therefore every grade must be equal or less than it
$72 \leqslant x \leqslant 95$
where is any quiz grade of any student in the class

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3 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is

Leto [7]

Answer:

$t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827$

The degrees of freedom are given by:

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And the p value taking in count that we have a bilateral test we got:

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Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value

Step-by-step explanation:

Information given

$\bar X=435$ represent the mean for the weight

$s=\sqrt{841}=29$ represent the sample standard deviation

$n=23$ sample size

$\mu_o =430$ represent the value that we want to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We are trying to proof if the filling machine works correctly at the 430 gram setting, so then the system of hypothesis for this case are:

Null hypothesis: $\mu = 430$

Alternative hypothesis: $\mu \neq 430$

In order to cehck the hypothesis the statistic for a one sample mean test is given by

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we have this:

$t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827$

The degrees of freedom are given by:

$df=n-1=23-1=22$

And the p value taking in count that we have a bilateral test we got:

$p_v =2*P(t_{(22)}>0.827)=0.417$

Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value

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3 years ago