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Murljashka [212]
3 years ago
13

i need you to increase the number of customers you talk to daily by 20%. I talk to an average of 8 customers per hour during an

8 hour shift so now i’ll need to talk to how many customers
Mathematics
1 answer:
Andre45 [30]3 years ago
7 0
First find out how many customers you talk to in an 8 hr shift
Rate is (8cs/hr)(8hr)=64 customers
So to increase by 20%, multiply the total number of customers by 120%, all the customers before plus the 20%, so (64)(1.2)=76.8, or 77 customers
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We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average sodium content.</em></u>

So, Null Hypothesis, H_0 : \mu = 1100 milligrams      {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, H_A : \mu \neq 1100 milligrams      {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about population standard deviation;

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

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            n = sample of packages of Oriental Spice Sauce = 40

So, <u><em>test statistics</em></u>  =  \frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}  ~ t_3_9

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The value of z test statistics is -0.307.

<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>

<em>Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the sodium content is same as what the nutrition label states.

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