A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289. Assume the population is normally distributed. A level of significance of 0.01 will be used. State the null and alternative hypotheses.

Null Hypothesis, : 430 gram   {means that the machine is not under filling the bags}

Alternate Hypothesis, : < 430 gram   {means that the machine is under filling the bags}

Step-by-step explanation:

We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting.

It is believed that the machine is under filling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289.

Let = mean weight bag filling capacity of machine.

SO, Null Hypothesis, : 430 gram   {means that the machine is not under filling the bags}

Alternate Hypothesis, : < 430 gram   {means that the machine is under filling the bags}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  =  ~

where,  = sample mean = 429 grams

             s = sample standard deviation = = 17 grams

             n = sample of bags = 21

So, test statistics  =    ~   

                               =  -0.269

Now at 0.01 significance level, the t table gives critical value of -2.528 at 20 degree of freedom for left-tailed test. Since our test statistics is higher than the critical value of t as -0.269 > -2.528, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the machine is not under filling the bags.

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)