A local fast-food restaurant serves buffalo wings. The restaurant's managers notice that they normally sell the following proportions of flavors for their wings: $30\mathrm{\%}$ Spicy Garlic, $20\mathrm{\%}$ Classic Medium, $20\mathrm{\%}$ Teriyaki, $10\mathrm{\%}$ Hot BBQ, and $20\mathrm{\%}$ Asian Zing. After running a campaign to promote their nontraditional specialty wings, they want to know if the campaign has made an impact. The results after 10 days are listed in the following table. Is there sufficient evidence at the 0.025 level of significance to say that the promotional campaign has made any difference in the proportions of flavors sold?
$$\text{Unknown environment 'tabular'}$$
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Step 1 of 4 : State the null and alternative hypotheses in terms of the expected proportion for each flavor.
$$\begin{array}{l}{H}_0:p_{\mathrm{SG}}=,p_{\mathrm{CM}}=\dots ,p_{\mathrm{BBQ}}=\\ p_{\mathrm{AZ}}=\\ H_a:\text{ There is a difference from the stated proportions. }\end{array}$$

Step 1 ：Step 1: State the null and alternative hypotheses. The null hypothesis $H_0$ is that the proportions of flavors sold remain the same as before the campaign, i.e., $p_{SG}=0.30$, $p_{CM}=0.20$, $p_T=0.20$, $p_{BBQ}=0.10$, and $p_{AZ}=0.20$. The alternative hypothesis $H_a$ is that there is a difference from the stated proportions.

Step 2 ：Step 2: Calculate the total number of wings sold. The total number of wings sold is $134+114+129+67=444$.

Step 3 ：Step 3: Calculate the observed proportions. The observed proportions are $p_{SG}=134/444=0.302$, $p_{CM}=114/444=0.257$, $p_T=129/444=0.291$, $p_{BBQ}=67/444=0.151$, and $p_{AZ}=0$.

Step 4 ：Step 4: Perform a chi-square test to compare the observed proportions with the expected proportions. The chi-square test statistic is ${\chi }^2=\sum _{i=1}^5\frac{(O_i-E_i{)}^2}{E_i}$, where $O_i$ is the observed frequency and $E_i$ is the expected frequency. The expected frequencies are $E_{SG}=444\ast 0.30=133.2$, $E_{CM}=444\ast 0.20=88.8$, $E_T=444\ast 0.20=88.8$, $E_{BBQ}=444\ast 0.10=44.4$, and $E_{AZ}=444\ast 0.20=88.8$. The observed frequencies are $O_{SG}=134$, $O_{CM}=114$, $O_T=129$, $O_{BBQ}=67$, and $O_{AZ}=0$. So, ${\chi }^2=\frac{(134-133.2{)}^2}{133.2}+\frac{(114-88.8{)}^2}{88.8}+\frac{(129-88.8{)}^2}{88.8}+\frac{(67-44.4{)}^2}{44.4}+\frac{(0-88.8{)}^2}{88.8}=1.21$.

Step 5 ：Step 5: Determine the critical value for the chi-square test at the 0.025 level of significance. The degrees of freedom for the test is $df=5-1=4$. The critical value for a chi-square test with 4 degrees of freedom at the 0.025 level of significance is 14.86.

Step 6 ：Step 6: Compare the test statistic with the critical value. Since the test statistic ${\chi }^2=1.21$ is less than the critical value 14.86, we do not reject the null hypothesis.

Step 7 ：Step 7: Conclusion. There is not sufficient evidence at the 0.025 level of significance to say that the promotional campaign has made any difference in the proportions of flavors sold. $\boxed{{\displaystyle \text{Final Answer: No significant difference.}}}$