Question

2. A brand of mints come in various flavors. The company says that it makes the mints in the following proportions.

Flavor	Cherry	Strawberry	Chocolate	Orange	Lime
Expected %	30%	20%	20%	15%	15%

A bag bought at random has the following number of mints in it.

Flavor	Cherry	Strawberry	Chocolate	Orange	Lime
Observed	67	50	54	29	25

Determine whether this distribution is consistent with company’s stated proportions.

1. What is the null hypothesis?
2. What is the alternative hypothesis?
3. Enter the observed number of times a flavor comes up in your test bag and the expected number of times that the flavor should come up into the X-squared goodness of fit applet.
4. What is the number of degrees of freedom?
5. What is the p-value? Provide a screen shot of your answer.
6. Using a 95% confidence interval, should you accept or reject the null hypothesis?
7. Does the distribution of flavors in your random bag support or contest the company’s state proportions? (yes or no).

Answer 1

The null hypothesis is H0 : The brand of mints come in various flavours are consistent with company's states proportions.

The alternative hypothesis is H1:The brand of mints come in various flavours are not consistent with company's states proportions.

We compute the table of observed frequencies and expected frequencies

here, expected frequencies gives in the % form that is in probability form (Pi)

We calculate expected frequencies = Ei = N*Pi

N =total observations = 225

Pi = proportion of each flavour (we convert it % into fraction values)

Observation table

The test statistic is given by

Oi = observed frequencies

Ei = expected frequencies

Flavour	Oi	Pi	Ei = N*Pi	Oi^2/Ei
Cherry	67	0.30	67.5	66.50
Strawberry	50	0.20	45	55.56
Chocolate	54	0.20	45	64.8
Orange	29	0.15	33.75	24.92
Lime	25	0.15	33.75	18.52
				= 230.3

test statistic is

............(1)

Degrees of freedom = n-k-1

n = total category of flavour = 5

k = number of independent observation = 0

Degrees of freedom = n-k-1 = 5-0-1 = 4

Degrees of freedom = 4

P-value

Oi	Ei
67	67.5
50	45
54	45
29	33.75
25	33.75

We find p-value in excel by using command

=CHITEST(actual range,expected_range)

select observed frequecies Oi in place of actual range and

expected frequencies in expected_range then press Enter key

P-value = 0.2582

We find 95% confidence interval and find the critical value of Chi-square by using excel is

=CHIINV(probability,degrees of freedom)

here 95% = 0.95 condidence interval

Critical value = CHIINV(0.05,4) then press Enter key in Excel

Critical value = 9.49.................(2)

By comparing calculated value in equation (1) with Critical value in equation (2)

and take decision of accept or reject null hypothesis

here,

Calculated value = 5.3 < Critical value = 9.49

So we should accept our null hypothesis.

that is the brand of mints come in various flavours are consistent with company's states proportions.