First let us fine mean and standard deviation
Create the following table.
data | data-mean | (data - mean)2 |
98.8 | 22.8833 | 523.64541889 |
50.3 | -25.6167 | 656.21531889 |
83.3 | 7.3833 | 54.51311889 |
67.3 | -8.6167 | 74.24751889 |
77.1 | 1.1833 | 1.40019889 |
78.7 | 2.7833 | 7.74675889 |
Find the sum of numbers in the last column to get.
So standard deviaiton is
Now let us assume that distribution is normal, so t value for 80% CI is TINV(0.2,5)=1.476
So Margin of Error is
Hence CI is
Calendar Gradebook essment e Intervals Due Sat 03/28/2020 11:00 pm You must estimate the mean temperature...
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I really need help with my Confidence Intervals for the Mean homework, can someone please help? 1. You measure 34 dogs' weights, and find they have a mean weight of 52 ounces. Assume the population standard deviation is 10.2 ounces. Based on this, construct a 99% confidence interval for the true population mean dog weight. Give your answers as decimals, to two places ________________ ± ______________ ounces 2. Assume that a sample is used to estimate a population mean μμ....
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Yifei Xie s Calendar Gradebook Sp18> Assessment work Due Thu 05/31/2018 11:59 pm You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a reasonable approximation for the binomial distribution in this case. while it is an uncommon confidence level, find the critical value that corresponds to a confidence level of97.5%. (Report answer accurate to three decimal places with appropriate rounding.) Points possible: 1 This is attempt 1 of 4. Submit
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