Question

Tropaeolum majus (nasturtium) is sold in bags at the farmers market. A patron is curious how many nasturtium will be in a 50 gram bag. In order to make an estimate the patron weighs ten random nasturtiums on a portable scale. The data is listed below in grams. Determine the C.I. for the true mean weight of an individual nasturtium at the 95% confidence level. Use the low and high values of the C.I. to determine the possible number of nasturtiums in a 50 gram bag. 1.45g, 1.35g, 1.45g, 1.65g, 1.72g, 1.45g, 1.35g, 1.92g, 1.76g, and 1.61g.

Answer 1

95% confidence interval for nasturtiums in a 50 gram bag is (1.442, 1.700)

Explanation:

The sample is:

1.45g, 1.35g, 1.45g, 1.65g, 1.72g, 1.45g, 1.35g, 1.92g, 1.76g, and 1.61g.

Sample size is 10

Mean nasturtiums in a 50 gram bag is
`(1.45g+1.35g+ 1.45g+ 1.65g+ 1.72g+ 1.45g+ 1.35g+ 1.92g+ 1.76g+1.61g )/(10) =1.571 g`

Standard deviation of the sample is the mean of the squared differences from the mean(0.181)

95% Confidence Interval can be calculated using M±ME where

• M is the mean nasturtiums in a 50 gram bag (1.571 g)

• ME is the margin of error from the mean

And margin of error (ME) can be calculated using the formula

ME=
`(t*s)/(√(N) )` where

• t is the corresponding statistic in the 95% confidence level and 9 degrees of freedom (2.262)

• s is the standard deviation of the sample (0.181)

• N is the sample size (10)

Then ME=
`(2.262*0.181)/(√(10) )` =0.129

95% confidence interval for nasturtiums in a 50 gram bag is 1.571±0.129