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Fiona grows large pumpkins and wants to estimate the mean weight of the approximately 500 pumpkins in her
harvest. She takes a random sample of 11 of these pumpkins and finds they have a mean weight of 44 kg and a
standard deviation of 6.7 kg. The weights in the sample were roughly symmetric with no outliers.

Sagot :

Answer:

44+-3.169(6.7/root11)

Step-by-step explanation:

The 99% confidence interval for the mean weight in kg of pumpkins grown by Fiona in the harvest for the provided sample is,

[tex]44\pm 3.169\times\dfrac{6.7}{\sqrt{11}}[/tex]

What is mean and standard deviation?

The mean of the data is the average value of the given data. The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

Fiona grows large pumpkins and wants to estimate the mean weight of the approximately 500 pumpkins in her harvest.

She takes a random sample of 11 of these pumpkins. Here, the sample size is 11 which is less than 30. Thus, use the formula mentioned below.

[tex]\overline x\pm t\times\dfrac{\sigma}{\sqrt{n}}[/tex]

Here, (n) is the sample size, [tex]\overline x[/tex] is the mean of the sample.

Degree of freedom is,

[tex]DF=11-1=10[/tex]

For this, the value of t from the t table is 3.169 for 99% confidence interval.

The mean weight of 44 kg and a standard deviation of 6.7 kg they find. Put these values in the above formula,

[tex]\overline x\pm t\times\dfrac{\sigma}{\sqrt{n}}=44\pm 3.169\times\dfrac{6.7}{\sqrt{11}}[/tex]

Thus, the 99% confidence interval for the mean weight in kg of pumpkins in the harvest for the provided sample is,

[tex]44\pm 3.169\times\dfrac{6.7}{\sqrt{11}}[/tex]

Learn more about the mean here;

https://brainly.com/question/14532771

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